## Precalculus (6th Edition)

$r=-\dfrac{h}{2}\pm\dfrac{\sqrt{\pi(\pi h^{2}+2S)}}{2\pi}$
$S=2\pi rh+2\pi r^{2}$ $,$ for $r$ Take $S$ to the right side of the equation and rearrange: $0=2\pi rh+2\pi r^{2}-S$ $2\pi r^{2}+2\pi hr-S=0$ Use the quadratic formula to solve the equation for $r$. The formula is $r=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=2\pi$, $b=2\pi h$ and $c=-S$ Substitute the known values into the formula and evaluate: $r=\dfrac{-2\pi h\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-S)}}{2(2\pi)}=...$ $...=\dfrac{-2\pi h\pm\sqrt{4\pi^{2}h^{2}+8\pi S}}{4\pi}=\dfrac{-2\pi h\pm\sqrt{4\pi(\pi h^{2}+2S)}}{4\pi}=...$ $...=\dfrac{-2\pi h\pm2\sqrt{\pi(\pi h^{2}+2S)}}{4\pi}=...$ $...=-\dfrac{h}{2}\pm\dfrac{\sqrt{\pi(\pi h^{2}+2S)}}{2\pi}$