# How do you solve 2p = sqrtp+5?

Set $p = {t}^{2} \ge 0$ hence we have that
2t^2=sqrt(t^2)+5=>2t^2=t+5=>2t^2-t-5=0=> t=1/4(1+sqrt41)
hence $p = {t}^{2} = \frac{1}{8} \left(21 + \sqrt{41}\right)$