Answer
$\left\{ \dfrac{-3}{5},1 \right\}$
Work Step by Step
The standard form of the given equation, $
2y=5y^2-3
,$ is
\begin{array}{l}\require{cancel}
-5y^2+2y+3=0
\\\\
5y^2-2y-3=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the given quadratic equation, $
5y^2-2y-3=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-2)\pm\sqrt{(-2)^2-4(5)(-3)}}{2(5)}
\\\\=
\dfrac{2\pm\sqrt{4+60}}{10}
\\\\=
\dfrac{2\pm\sqrt{64}}{10}
\\\\=
\dfrac{2\pm\sqrt{(8)^2}}{10}
\\\\=
\dfrac{2\pm8}{10}
\\\\=
\dfrac{2-8}{10}
\text{ OR }
\dfrac{2+8}{10}
\\\\=
\dfrac{-6}{10}
\text{ OR }
\dfrac{10}{10}
\\\\=
\dfrac{-3}{5}
\text{ OR }
1
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{-3}{5},1 \right\}
.$
