Answer
$\left\{ \dfrac{-1-i\sqrt{23}}{4},\dfrac{-1+i\sqrt{23}}{4} \right\}$
Work Step by Step
The standard form of the given equation, $
\dfrac{2}{5}y^2+\dfrac{1}{5}y+\dfrac{3}{5}=0
,$ is
\begin{array}{l}\require{cancel}
5\left( \dfrac{2}{5}y^2+\dfrac{1}{5}y+\dfrac{3}{5} \right)=(0)5
\\\\
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{-23}}{4}
\\\\=
\dfrac{-1\pm i\sqrt{23}}{4}
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{-1-i\sqrt{23}}{4},\dfrac{-1+i\sqrt{23}}{4} \right\}
.$
