Answer
$\left\{ -1,\dfrac{3}{2} \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{2}{5}y^2+\dfrac{1}{5}y=\dfrac{3}{5}
,$ is
\begin{array}{l}\require{cancel}
5\left( \dfrac{2}{5}y^2+\dfrac{1}{5}y \right)=\left( \dfrac{3}{5} \right)5
\\\\
1(2y^2)+1(y)=3(1)
\\\\
2y^2+y=3
\\\\
2y^2+y-3=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
2y^2+y-3=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-1)\pm\sqrt{(-1)^2-4(2)(-3)}}{2(2)}
\\\\=
\dfrac{1\pm\sqrt{1+24}}{4}
\\\\=
\dfrac{1\pm\sqrt{25}}{4}
\\\\=
\dfrac{1\pm\sqrt{(5)^2}}{4}
\\\\=
\dfrac{1\pm5}{4}
\\\\=
\dfrac{1-5}{4}
\text{ OR }
\dfrac{1+5}{4}
\\\\=
\dfrac{-4}{4}
\text{ OR }
\dfrac{6}{4}
\\\\=
-1
\text{ OR }
\dfrac{3}{2}
.\end{array}
Hence, the solutions are $
\left\{ -1,\dfrac{3}{2} \right\}
.$
