Answer
$1 \text{ }$
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
\\\\
1(y^2)=y(2)-1(1)
\\\\
y^2=2y-1
\\\\
y^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, $
y^2-2y+1=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(1)}}{2(1)}
\\\\=
\dfrac{2\pm\sqrt{4-4}}{2}
\\\\=
\dfrac{2\pm\sqrt{0}}{2}
\\\\=
\dfrac{2\pm0}{2}
\\\\=
\dfrac{2}{2}
\\\\=
1
.\end{array}
Hence, the solutions are $
1 \text{ }
.$
