Answer
$\displaystyle\frac{-\pi h\pm\sqrt{\pi^{2}h^{2}+2\pi A}}{2\pi}$
Work Step by Step
$A = 2\pi r^{2}+2\pi rh$
$2\pi r^{2}+2\pi hr-A = 0$
We use the quadratic formula:
$r=\displaystyle \frac{-(2\pi h)\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-A)}}{2(2\pi)}=\frac{-2\pi h\pm\sqrt{4\pi^{2}h^{2}+8\pi A}}{4\pi}=\frac{-2\pi h\pm2\sqrt{\pi^{2}h^{2}+2\pi A}}{4\pi}=\frac{-\pi h\pm\sqrt{\pi^{2}h^{2}+2\pi A}}{2\pi}$
