Originally Posted by

**masters** This is a monster. Where did this come from?

$\displaystyle pr^2+2prR+pR^2=E^2R$

$\displaystyle pr^2+2prR+pR^2-E^2R=0$

$\displaystyle pR^2+2prR-E^2R=-pr^2$

$\displaystyle pR^2+(2pr-E^2)R=-pr^2$

Divide everything by p.

$\displaystyle R^2+\left(\frac{2pr-E^2}{p}\right)R=-r^2$

Try completing the square.

$\displaystyle R^2+\left(\frac{2pr-E^2}{p}\right)R+\left(\frac{2pr-E^2}{2p}\right)^2=-r^2+\left(\frac{2pr-E^2}{2p}\right)^2$

$\displaystyle \left(R+\frac{2pr-E^2}{2p}\right)^2=-r^2+\left(\frac{2pr-E^2}{2p}\right)^2$

Take the square root of both sides.

$\displaystyle R+\frac{2pr-E}{2p}=\pm\sqrt{-r^2+\left(\frac{2pr-E}{2p}\right)^2}$

$\displaystyle R=\pm\sqrt{-r^2+\left(\frac{2pr-E}{2p}\right)^2}-\frac{2pr-E}{2p}$

I don't know if the right side gets any better. Maybe an algebraist will take a look at this and submit a cleaner solution.