<s>[INST] Solve for $+r$ ; $A=2\pi r^2+2\pi rh$ 
Solve for $+r$

$$A=2\pi r^2+2\pi rh$$ Since $2\pi$ is common on both sides of the $+$ so I will take it out
$$A=2\pi (r^2+rh)$$ 
Now, divide both sides by $2\pi$
$$\dfrac{A}{2\pi}=r^2+rh$$
Then, we can divide by the $h$
$$\dfrac{Ah}{2\pi}=r^2+r$$
Then; 
$$0=r^2+r-\dfrac{Ah}{2\pi}$$ 
Quadratic formula $=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=1,b=1,c=?$ What are the values of $a,b,c$
 [/INST] There is a series of mistakes in your solution. I'll mark them one by one:
You correctly arrive to 
$$\frac A {2\pi}=r^2+rh$$
But "dividing by $h$" produces
$$\frac 1 h \frac A {2\pi}=\frac 1 h\left(r^2+rh\right)$$
$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+\frac{rh}h$$
$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+r$$
So that step is wrong.
Similarily, if you have 
$$\dfrac{Ah}{2\pi}=r^2+r$$
then "taking square roots" produces
$$\sqrt{\dfrac{Ah}{2\pi}}=\sqrt{r^2+r}$$
You then seem to assert
$$\sqrt{r^2+r}=2r$$
Let's check if it is indeed true for, say $r=1$, which gives
$$\sqrt{2}=2$$
which is manifestly false. So there is something awry there, too.
The best thing you can do is check wether each step is correct. To solve for $r$, since
$$\frac A {2\pi}=r^2+rh$$
is a quadratic we need to make a very old trick, which is called completing the square:
$$\eqalign{
  & \frac{A}{{2\pi }} = {r^2} + rh  \cr 
  & \frac{A}{{2\pi }} = {r^2} + 2r\frac{h}{2}  \cr 
  & \frac{A}{{2\pi }} = \underbrace {{r^2} + 2r\frac{h}{2} + {{\left( {\frac{h}{2}} \right)}^2}}_{{\text{This is a perfect square!}}} - {\left( {\frac{h}{2}} \right)^2}  \cr 
  & \frac{A}{{2\pi }} = {\left( {r + \frac{h}{2}} \right)^2} - {\left( {\frac{h}{2}} \right)^2}  \cr 
  & \frac{A}{{2\pi }} + {\left( {\frac{h}{2}} \right)^2} = {\left( {r + \frac{h}{2}} \right)^2}  \cr 
  & \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}}  = {{{\left( {r + \frac{h}{2}} \right)}^2}}   \cr 
  & \pm \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}}  =  {r + \frac{h}{2}}  \cr} $$
Note in the last steps we take the square root. We then have to think about both the positive and negative root. So you final solution is 
$$r =  - \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}}  - \frac{h}{2}{\text{ or }}r = \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}}  - \frac{h}{2}$$
COMPLETING THE SQUARE:
Say we have a quadratic $$0=ax^2+bx+c$$
"Completing the square" consist of writing it in the form 
$$0 = A{\left( {x + h} \right)^2} + C$$
We can accomplish this with some "trickery"
$$\eqalign{
  & 0 = a{x^2} + bx + c  \cr 
  & 0 = 4{a^2}{x^2} + 4abx + 4ac{\text{ ;  multiply by }}4a  \cr 
  & 0 = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + 4ac{\text{ ; cleverly rearrange the eqn}}{\text{.}}  \cr 
  & {b^2} = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2} + 4ac{\text{ ;  add }}{b^2}  \cr 
  & {b^2} = \underbrace {{{\left( {2ax} \right)}^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2}}_{{\text{This is a perfect square!}}} + 4ac  \cr 
  & {b^2} = {\left( {2ax + b} \right)^2} + 4ac  \cr 
  & {b^2} - 4ac = {\left( {2ax + b} \right)^2}  \cr 
  & \sqrt {{b^2} - 4ac}  = 2ax + b  \cr 
  &  - b + \sqrt {{b^2} - 4ac}  = 2ax  \cr 
  & \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = x \cr} $$
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