Answer
$R=\pm\dfrac{\sqrt{Vh-r^2h}}{h}$
Work Step by Step
Using the properties of equality, in terms of $
R
$, the given equation, $
V=r^2+R^2h
,$ is equivalent to
\begin{align*}\require{cancel}
V-r^2&=r^2+R^2h-r^2
\\
V-r^2&=R^2h
\\\\
\dfrac{V-r^2}{h}&=\dfrac{R^2\cancel h}{\cancel h}
\\\\
\dfrac{V-r^2}{h}&=R^2
\\\\
R^2&=\dfrac{V-r^2}{h}
.\end{align*}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{align*}
R&=\pm\sqrt{\dfrac{V-r^2}{h}}
.\end{align*}
Using the properties of radicals, the equation above is equivalent to
\begin{align*}
R&=\pm\sqrt{\dfrac{V-r^2}{h}\cdot\dfrac{h}{h}}
&(\text{rationalize the denominator})
\\\\
R&=\pm\sqrt{\dfrac{1}{h^2}\cdot (Vh-r^2h)}
\\\\
R&=\pm\sqrt{\dfrac{1}{h^2}}\cdot \sqrt{Vh-r^2h}
\\\\
R&=\pm\dfrac{\sqrt{Vh-r^2h}}{h}
.\end{align*}
Hence, $
V=r^2+R^2h
$ is equivalent to
\begin{align*}
R=\pm\dfrac{\sqrt{Vh-r^2h}}{h}
.\end{align*}
