Answer
$R=\dfrac{E^2-2pr\pm E\sqrt{E^2-4pr}}{2p}$
Work Step by Step
Using the properties of equality, in terms of $R$, the given equation, $
p=\dfrac{E^2R}{(r+R)^2}
,$ is equivalent to
\begin{align*}\require{cancel}
(r+R)^2(p)&=\dfrac{E^2R}{\cancel{(r+R)^2}}\cdot\cancel{(r+R)^2}
\\\\
(r+R)^2(p)&=E^2R
\\
\left[(r)^2+2(r)(R)+(R)^2\right](p)&=E^2R
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\
\left[r^2+2rR+R^2\right](p)&=E^2R
\\
r^2p+2rRp+R^2p&=E^2R
.\end{align*}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{align*}
r^2p+2rRp+R^2p-E^2R&=0
\\
R^2p+(2rRp-E^2R)+r^2p&=0
\\
(p)R^2+(2rp-E^2)R+r^2p&=0
.\end{align*}
The equation above has
\begin{align*}
a=
p
,\text{ }b=
(2rp-E^2)
,\text{ and }c=
r^2p
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
R&=
\dfrac{-(2rp-E^2)\pm\sqrt{(2rp-E^2)^2-4(p)(r^2p)}}{2(p)}
\\\\&=
\dfrac{-2rp+E^2\pm\sqrt{\left[(2rp)^2-2(2rp)(E^2)+(E^2)^2\right]-4r^2p^2}}{2p}
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\\\&=
\dfrac{-2rp+E^2\pm\sqrt{4r^2p^2-4rpE^2+E^4-4r^2p^2}}{2p}
\\\\&=
\dfrac{-2rp+E^2\pm\sqrt{\cancel{4r^2p^2}-4rpE^2+E^4-\cancel{4r^2p^2}}}{2p}
\\\\&=
\dfrac{-2rp+E^2\pm\sqrt{E^4-4rpE^2}}{2p}
\\\\&=
\dfrac{-2rp+E^2\pm\sqrt{E^2(E^2-4rp)}}{2p}
\\\\&=
\dfrac{-2rp+E^2\pm E\sqrt{E^2-4rp}}{2p}
\\\\&=
\dfrac{E^2-2pr\pm E\sqrt{E^2-4pr}}{2p}
.\end{align*}
Hence, in terms of $R,$ the given equation is equivalent to $R=\dfrac{E^2-2pr\pm E\sqrt{E^2-4pr}}{2p}$.
