Surface area of a cylinder- finding unknown variable

[opus]

Homework Statement

The surface area, A, of a cylinder with height, h, and radius, r, is given by the equation $A=2πrh+2πr^2$.
A company makes soup cans by using 32π square inches of aluminum sheet for each can. If the height of the can is 6 inches, find the radius of the can.

Homework Equations

$A=2πrh+2πr^2$

The Attempt at a Solution

$A=32πin^2$
$h= 6in$
I start by recognizing that I need to find the value of r in the given equation.
Next, I plug the known values into the equation, which gives me $32π in^2=2πr\left(6in\right)+2πr^2$
I don't know how to go about isolating r, considering there is an r and an $r^2$ term.
In my attempt, I came up with $\left( \frac {32πin^2} {2π6} \right) = r + 2πr^2$ and it seems to not be the right idea, because I feel like next I would want to divide the RHS by 2π, leaving me with a complex fraction on the left, and an $r+r^2$ on the right.
Any tips to get me going in the right direction?

 

[fresh_42]

Homework Statement

The surface area, A, of a cylinder with height, h, and radius, r, is given by the equation $A=2πrh+2πr^2$.
A company makes soup cans by using 32π square inches of aluminum sheet for each can. If the height of the can is 6 inches, find the radius of the can.

Homework Equations

$A=2πrh+2πr^2$

The Attempt at a Solution

$A=32πin^2$
$h= 6in$
I start by recognizing that I need to find the value of r in the given equation.
Next, I plug the known values into the equation, which gives me $32π in^2=2πr\left(6in\right)+2πr^2$
I don't know how to go about isolating r, considering there is an r and an $r^2$ term.
In my attempt, I came up with $\left( \frac {32πin^2} {2π6} \right) = r + 2πr^2$ and it seems to not be the right idea, because I feel like next I would want to divide the RHS by 2π, leaving me with a complex fraction on the left, and an $r+r^2$ on the right.
Any tips to get me going in the right direction?

What do you get, if you divide your equation by $2\pi$ and write it as $0=r^2 + \ldots \,$?
If you still have no idea then, write it with $x=r$ with $x$ instead of $r$.

 

[opus]

Wunderbar! Kind of a tricky one.
$r^2+6r-\left(\frac {32πin^2} {2π} \right) = 0$
Which gives me
r=-8 in or r=2 in
Since r cannot be negative, it is 2 inches.

Thank you fresh_42