Age problem-PLEASE HELP [closed]
$begingroup$
Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!
quadratics
edited Jan 4 at 6:42
Sauhard Sharma
953318
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
$endgroup$
closed as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!
quadratics
edited Jan 4 at 6:42
Sauhard Sharma
953318
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
$endgroup$
closed as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
It's a quadratic equation. Solve it.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
Quadratic formula?
$endgroup$
– Thomas Shelby
Jan 4 at 6:28
2
$begingroup$
Also this is not a linear-algebra problem.
$endgroup$
– harshit54
Jan 4 at 6:28
1
$begingroup$
Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
$endgroup$
– JMoravitz
Jan 4 at 6:30
$begingroup$
As an aside, I get $32x^2+4x-1176=0$ as the equation with the constant term as $-1176$, not $-1179$. You might try doublechecking your arithmetic.
$endgroup$
– JMoravitz
Jan 4 at 6:33
add a comment |
$begingroup$
Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!
quadratics
edited Jan 4 at 6:42
Sauhard Sharma
953318
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
$endgroup$
Hannah. added one to her age, then multiplied the result by $4$. She squared that product, then increased the result by $3$, then doubled it. The final result was $1214$ more than $60$ times her age. Find Hannah.'s age, which we can assume is a positive number. I simplified to $32x^2+4x-1179=0$ but I don't really know what to do now. Please help!
quadratics
quadratics
edited Jan 4 at 6:42
Sauhard Sharma
953318
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
edited Jan 4 at 6:42
Sauhard Sharma
953318
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
edited Jan 4 at 6:42
Sauhard Sharma
953318
edited Jan 4 at 6:42
Sauhard Sharma
953318
edited Jan 4 at 6:42
Sauhard Sharma
953318
953318
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
asked Jan 4 at 6:25
J. DOEEJ. DOEE
16927
16927
closed as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo Jan 4 at 12:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Xander Henderson, Nosrati, Paul Frost, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
It's a quadratic equation. Solve it.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
Quadratic formula?
$endgroup$
– Thomas Shelby
Jan 4 at 6:28
2
$begingroup$
Also this is not a linear-algebra problem.
$endgroup$
– harshit54
Jan 4 at 6:28
1
$begingroup$
Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
$endgroup$
– JMoravitz
Jan 4 at 6:30
$begingroup$
As an aside, I get $32x^2+4x-1176=0$ as the equation with the constant term as $-1176$, not $-1179$. You might try doublechecking your arithmetic.
$endgroup$
– JMoravitz
Jan 4 at 6:33
add a comment |
$begingroup$
It's a quadratic equation. Solve it.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
Quadratic formula?
$endgroup$
– Thomas Shelby
Jan 4 at 6:28
2
$begingroup$
Also this is not a linear-algebra problem.
$endgroup$
– harshit54
Jan 4 at 6:28
1
$begingroup$
Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
$endgroup$
– JMoravitz
Jan 4 at 6:30
$begingroup$
As an aside, I get $32x^2+4x-1176=0$ as the equation with the constant term as $-1176$, not $-1179$. You might try doublechecking your arithmetic.
$endgroup$
– JMoravitz
Jan 4 at 6:33
$begingroup$
It's a quadratic equation. Solve it.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
It's a quadratic equation. Solve it.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
Quadratic formula?
$endgroup$
– Thomas Shelby
Jan 4 at 6:28
$begingroup$
Quadratic formula?
$endgroup$
– Thomas Shelby
Jan 4 at 6:28
2
2
$begingroup$
Also this is not a linear-algebra problem.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
Also this is not a linear-algebra problem.
$endgroup$
– harshit54
Jan 4 at 6:28
1
1
$begingroup$
Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
$endgroup$
– JMoravitz
Jan 4 at 6:30
$begingroup$
Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
$endgroup$
– JMoravitz
Jan 4 at 6:30
$begingroup$
As an aside, I get $32x^2+4x-1176=0$ as the equation with the constant term as $-1176$, not $-1179$. You might try doublechecking your arithmetic.
$endgroup$
– JMoravitz
Jan 4 at 6:33
$begingroup$
As an aside, I get $32x^2+4x-1176=0$ as the equation with the constant term as $-1176$, not $-1179$. You might try doublechecking your arithmetic.
$endgroup$
– JMoravitz
Jan 4 at 6:33
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Letting Hannah's age be $x$ we obtain that
$$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$
$$32x^2+4x-1176=0$$
Using quadratic formula, we obtain $x=6$ (taking the positive result only).
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
$endgroup$
add a comment |
$begingroup$
Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
$$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
$endgroup$
add a comment |
$begingroup$
Your equation is slightly incorrect. The actual equation is as follows(based on your question)
$$32x^2 +4x -1176=0$$
This equation can be factored using the quadratic formula
$$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$
$$x=frac{-4pm sqrt{150544}}{64}$$
$$x=frac{-4pm 388}{64}$$
Since age can't be negative
$$x=(-4+388)/64 = 6$$
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
$32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.
The $ac$ method requires us to find two integers whose product is
$8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.
We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are
$$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$
Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.
begin{align}
32x^2 +4x -1176
&= 4(8x^2 + color{red}1x - 294) \
&= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
&= 4((8x^2 - 48x) + (49x - 294)) \
&= 4((8x(x - 6) + 49(x - 6)) \
&= 4(8x+49)(x - 6) \
end{align}
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Letting Hannah's age be $x$ we obtain that
$$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$
$$32x^2+4x-1176=0$$
Using quadratic formula, we obtain $x=6$ (taking the positive result only).
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
$endgroup$
add a comment |
$begingroup$
Letting Hannah's age be $x$ we obtain that
$$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$
$$32x^2+4x-1176=0$$
Using quadratic formula, we obtain $x=6$ (taking the positive result only).
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
$endgroup$
add a comment |
$begingroup$
Letting Hannah's age be $x$ we obtain that
$$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$
$$32x^2+4x-1176=0$$
Using quadratic formula, we obtain $x=6$ (taking the positive result only).
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
$endgroup$
Letting Hannah's age be $x$ we obtain that
$$2left( left(4(x+1)right)^2 +3 right)-1214=60x$$
$$32x^2+4x-1176=0$$
Using quadratic formula, we obtain $x=6$ (taking the positive result only).
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
answered Jan 4 at 6:35
Hugh EntwistleHugh Entwistle
846217
846217
add a comment |
add a comment |
$begingroup$
Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
$$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
$endgroup$
add a comment |
$begingroup$
Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
$$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
$endgroup$
add a comment |
$begingroup$
Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
$$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
$endgroup$
Assuming that your result is correct, we are left with the quadratic equation $32x^2+4x-1179=0$. To solve for $x$, there are various techniques. The most common (and general) one would be completing the square, which will lead you to the general quadratic formula, allowing you to solve any quadratic equation of the form $ax^2+bx+c=0$ by
$$x=frac1{2a}left(-bpmsqrt{b^2-4ac}right).$$
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
answered Jan 4 at 6:34
YiFanYiFan
5,1052727
5,1052727
add a comment |
add a comment |
$begingroup$
Your equation is slightly incorrect. The actual equation is as follows(based on your question)
$$32x^2 +4x -1176=0$$
This equation can be factored using the quadratic formula
$$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$
$$x=frac{-4pm sqrt{150544}}{64}$$
$$x=frac{-4pm 388}{64}$$
Since age can't be negative
$$x=(-4+388)/64 = 6$$
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
Your equation is slightly incorrect. The actual equation is as follows(based on your question)
$$32x^2 +4x -1176=0$$
This equation can be factored using the quadratic formula
$$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$
$$x=frac{-4pm sqrt{150544}}{64}$$
$$x=frac{-4pm 388}{64}$$
Since age can't be negative
$$x=(-4+388)/64 = 6$$
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
Your equation is slightly incorrect. The actual equation is as follows(based on your question)
$$32x^2 +4x -1176=0$$
This equation can be factored using the quadratic formula
$$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$
$$x=frac{-4pm sqrt{150544}}{64}$$
$$x=frac{-4pm 388}{64}$$
Since age can't be negative
$$x=(-4+388)/64 = 6$$
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
$endgroup$
Your equation is slightly incorrect. The actual equation is as follows(based on your question)
$$32x^2 +4x -1176=0$$
This equation can be factored using the quadratic formula
$$x=frac{-4pm sqrt{16+4cdot1176cdot32}}{2cdot32}$$
$$x=frac{-4pm sqrt{150544}}{64}$$
$$x=frac{-4pm 388}{64}$$
Since age can't be negative
$$x=(-4+388)/64 = 6$$
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
answered Jan 4 at 6:34
Sauhard SharmaSauhard Sharma
953318
953318
add a comment |
add a comment |
$begingroup$
$32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.
The $ac$ method requires us to find two integers whose product is
$8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.
We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are
$$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$
Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.
begin{align}
32x^2 +4x -1176
&= 4(8x^2 + color{red}1x - 294) \
&= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
&= 4((8x^2 - 48x) + (49x - 294)) \
&= 4((8x(x - 6) + 49(x - 6)) \
&= 4(8x+49)(x - 6) \
end{align}
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
$32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.
The $ac$ method requires us to find two integers whose product is
$8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.
We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are
$$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$
Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.
begin{align}
32x^2 +4x -1176
&= 4(8x^2 + color{red}1x - 294) \
&= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
&= 4((8x^2 - 48x) + (49x - 294)) \
&= 4((8x(x - 6) + 49(x - 6)) \
&= 4(8x+49)(x - 6) \
end{align}
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
$32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.
The $ac$ method requires us to find two integers whose product is
$8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.
We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are
$$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$
Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.
begin{align}
32x^2 +4x -1176
&= 4(8x^2 + color{red}1x - 294) \
&= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
&= 4((8x^2 - 48x) + (49x - 294)) \
&= 4((8x(x - 6) + 49(x - 6)) \
&= 4(8x+49)(x - 6) \
end{align}
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
$endgroup$
$32x^2 +4x -1176=0$ can be divided by $4$, resulting in $8x^2 + x - 294 = 0$.
The $ac$ method requires us to find two integers whose product is
$8 cdot (-294) = -2cdot2cdot 2 cdot 2 cdot 3 cdot 7 cdot 7$ and whose sum is $1$.
We note that $2$ cannot be a factor of both integers since then their sum will be a multiple of $2$ and $1$ is not a multiple of $2$. Likewise, 7 cannot be a factor of both integers. This leads us to guess that the two integers are
$$text{$-2cdot 2 cdot 2 cdot 2 cdot 3 = -48 qquad $ and $qquad 7 cdot 7 = 49$}$$
Next we replace $x =1x$ with $1x = -48x + 49x$ and factor.
begin{align}
32x^2 +4x -1176
&= 4(8x^2 + color{red}1x - 294) \
&= 4(8x^2 - color{red}{48}x + color{red}{49}x - 294) \
&= 4((8x^2 - 48x) + (49x - 294)) \
&= 4((8x(x - 6) + 49(x - 6)) \
&= 4(8x+49)(x - 6) \
end{align}
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
answered Jan 4 at 8:12
steven gregorysteven gregory
18.3k32358
18.3k32358
add a comment |
add a comment |
$begingroup$
It's a quadratic equation. Solve it.
$endgroup$
– harshit54
Jan 4 at 6:28
$begingroup$
Quadratic formula?
$endgroup$
– Thomas Shelby
Jan 4 at 6:28
2
$begingroup$
Also this is not a linear-algebra problem.
$endgroup$
– harshit54
Jan 4 at 6:28
1
$begingroup$
Assuming your work done so far was correct, you are left with an equation of the form $Ax^2+ Bx + C = 0$, to which the possible values for $x$ would be $dfrac{-B+sqrt{B^2-4AC}}{2A}$ and $dfrac{-B-sqrt{B^2-4AC}}{2A}$. Find those values by plugging in the appropriate values for $A,B,C$ and note that people cannot be a negative number of years old (and still be capable of arithmetic).
$endgroup$
– JMoravitz
Jan 4 at 6:30
$begingroup$
As an aside, I get $32x^2+4x-1176=0$ as the equation with the constant term as $-1176$, not $-1179$. You might try doublechecking your arithmetic.
$endgroup$
– JMoravitz
Jan 4 at 6:33