Question 1160931
{{{P(R+b)^2=RE}}}

{{{P(R^2+2Rb+b^2)=RE}}}

{{{PR^2+2PbR+Pb^2=RE}}}

{{{PR^2+2PbR-ER+Pb^2=0}}}

{{{PR^2+(2Pb-E)R+Pb^2=0}}}

Use the general solution formula for quadratic equation, to solve for R.