Answer
$\left\{ \dfrac{1-\sqrt{3}}{2},\dfrac{1+\sqrt{3}}{2} \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{1}{2}y^2=y+\dfrac{1}{2}
,$ is
\begin{array}{l}\require{cancel}
2\left( \dfrac{1}{2}y^2 \right)=\left( y+\dfrac{1}{2} \right)2
\\\\
2(y^2)=y(2)+1(1)
\\\\
2y^2=2y+1
\\\\
2y^2-2y-1=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
2y^2-2y-1=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}
\\\\=
\dfrac{2\pm\sqrt{4+8}}{4}
\\\\=
\dfrac{2\pm\sqrt{12}}{4}
\\\\=
\dfrac{2\pm\sqrt{4\cdot3}}{4}
\\\\=
\dfrac{2\pm\sqrt{(2)^2\cdot3}}{4}
\\\\=
\dfrac{2\pm2\sqrt{3}}{4}
\\\\=
\dfrac{2(1\pm\sqrt{3})}{4}
\\\\=
\dfrac{\cancel{2}(1\pm\sqrt{3})}{\cancel{2}\cdot2}
\\\\=
\dfrac{1\pm\sqrt{3}}{2}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{1-\sqrt{3}}{2},\dfrac{1+\sqrt{3}}{2} \right\}
.$
