Answer
$r=\dfrac{-\pi h\pm\sqrt{\pi^2 h^2+\pi S}}{\pi}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
S=2\pi rh+\pi r^2
,$ is equivalent to
\begin{align*}\require{cancel}
0&=2\pi rh+\pi r^2-S
\\
0&=\pi r^2+2\pi rh-S
\\
0&=(\pi) r^2+(2\pi h) r-S
.\end{align*}
The equation above has
\begin{align*}
a=
\pi
,\text{ }b=
2\pi h
,\text{ and }c=
-S
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
r&=
\dfrac{-(2\pi h)\pm\sqrt{(2\pi h)^2-4(\pi)(-S)}}{2(\pi)}
\\\\&=
\dfrac{-2\pi h\pm\sqrt{4\pi^2 h^2+4\pi S}}{2\pi}
\\\\&=
\dfrac{-2\pi h\pm\sqrt{4\cdot(\pi^2 h^2+\pi S)}}{2\pi}
\\\\&=
\dfrac{-2\pi h\pm\sqrt{4}\cdot\sqrt{\pi^2 h^2+\pi S}}{2\pi}
\\\\&=
\dfrac{-2\pi h\pm2\sqrt{\pi^2 h^2+\pi S}}{2\pi}
\\\\&=
\dfrac{-\cancel2\pi h\pm\cancel2\sqrt{\pi^2 h^2+\pi S}}{\cancel2\pi}
\\\\&=
\dfrac{-\pi h\pm\sqrt{\pi^2 h^2+\pi S}}{\pi}
.\end{align*}
Hence, in terms of $r,$ the given equation is equivalent to $r=\dfrac{-\pi h\pm\sqrt{\pi^2 h^2+\pi S}}{\pi}$.
