#### Answer

$p=\dfrac{3}{4}\pm\dfrac{\sqrt{5}}{4}$

#### Work Step by Step

$\Big(p-\dfrac{1}{2}\Big)^{2}=\dfrac{p}{2}$
Evaluate the power on the left side of the equation:
$p^{2}-p+\dfrac{1}{4}=\dfrac{p}{2}$
Multiply the whole equation by $4$ to avoid working with fractions:
$4\Big(p^{2}-p+\dfrac{1}{4}=\dfrac{p}{2}\Big)$
$4p^{2}-4p+1=2p$
Take all terms to the left side:
$4p^{2}-4p-2p+1=0$
Simplify the equation by combining like terms:
$4p^{2}-6p+1=0$
Use the quadratic formula to solve this equation. The formula is $p=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=4$, $b=-6$ and $c=1$.
Substitute:
$p=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(4)(1)}}{2(4)}=\dfrac{6\pm\sqrt{36-16}}{8}=...$
$...=\dfrac{6\pm\sqrt{20}}{8}=\dfrac{6\pm2\sqrt{5}}{8}=\dfrac{3}{4}\pm\dfrac{\sqrt{5}}{4}$