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} $