College Algebra, Chapter 1, 1.3, Section 1.3, Problem 62

Solve $A = 2 \pi r^2 + 2 \pi rh$ for $r$.


$
\begin{equation}
\begin{aligned}

A =& 2 \pi r^2 + 2 \pi rh
&& \text{Given}
\\
\\
A =& 2 \pi (r^2 + rh)
&& \text{Factor out } 2 \pi
\\
\\
\frac{A}{2 \pi} =& r^2 + rh
&& \text{Divide both sides by } 2 \pi
\\
\\
\frac{A}{2 \pi} + \frac{h^2}{4} =& r^2 + rh + \frac{h^2}{4}
&& \text{Complete the square: add } \left( \frac{h}{2} \right)^2 = \frac{h^2}{4}
\\
\\
\frac{A}{2 \pi} + \frac{h^2}{4} =& \left( r + \frac{h}{2} \right)^2
&& \text{Perfect square}
\\
\\
\pm \sqrt{\frac{2A + h^2}{4 \pi}} =& r + \frac{h}{2}
&& \text{Take the square root and simplify by using LCD}
\\
\\
r =& \frac{-h}{2} \pm \sqrt{\frac{2A + h^2}{4 \pi}}
&& \text{Subtract } \frac{h}{2}
\\
\\
r =& \frac{-h}{2} \pm \frac{1}{2} \left( \sqrt{\frac{2A + h^2}{\pi}} \right)
&& \text{Solve for } r



\end{aligned}
\end{equation}
$