Answer
$\left\{ \dfrac{3-\sqrt{5}}{4},\dfrac{3+\sqrt{5}}{4} \right\}$
Work Step by Step
The standard form of the given equation, $
\left( p-\dfrac{1}{2} \right)^2=\dfrac{p}{2}
,$ is
\begin{array}{l}\require{cancel}
(p)^2+2(p)\left(-\dfrac{1}{2}\right)+\left(-\dfrac{1}{2}\right)^2=\dfrac{p}{2}
\\\\
p^2-p+\dfrac{1}{4}=\dfrac{p}{2}
\\\\
4\left( p^2-p+\dfrac{1}{4} \right)=\left(\dfrac{p}{2}\right)(4)
\\\\
4(p^2)+4(-p)+1(1)=p(2)
\\\\
4p^2-4p+1=2p
\\\\
4p^2+(-4p-2p)+1=0
\\\\
4p^2-6p+1=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
4p^2-6p+1=0
,$ are
\begin{array}{l}\require{cancel}
\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\pm\sqrt{4\cdot5}}{8}
\\\\=
\dfrac{6\pm\sqrt{(2)^2\cdot5}}{8}
\\\\=
\dfrac{6\pm2\sqrt{5}}{8}
\\\\=
\dfrac{2(3\pm\sqrt{5})}{8}
\\\\=
\dfrac{\cancel{2}(3\pm\sqrt{5})}{\cancel{2}\cdot4}
\\\\=
\dfrac{3\pm\sqrt{5}}{4}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{3-\sqrt{5}}{4},\dfrac{3+\sqrt{5}}{4} \right\}
.$
