#### Answer

$y=-\dfrac{1}{4}\pm\dfrac{\sqrt{23}}{4}i$

#### Work Step by Step

$\dfrac{2}{5}y^{2}+\dfrac{1}{5}y+\dfrac{3}{5}=0$
Multiply the whole equation by $5$ to avoid working with fractions:
$5\Big(\dfrac{2}{5}y^{2}+\dfrac{1}{5}y+\dfrac{3}{5}=0\Big)$
$2y^{2}+y+3=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=2$, $b=1$ and $c=3$.
Substitute:
$x=\dfrac{-1\pm\sqrt{1^{2}-4(2)(3)}}{2(2)}=\dfrac{-1\pm\sqrt{1-24}}{4}=...$
$...=\dfrac{-1\pm\sqrt{-23}}{4}=\dfrac{-1\pm\sqrt{23}i}{4}=-\dfrac{1}{4}\pm\dfrac{\sqrt{23}}{4}i$