#### Answer

$y=1$ and $y=-\dfrac{3}{2}$

#### Work Step by Step

$\dfrac{2}{5}y^{2}+\dfrac{1}{5}y=\dfrac{3}{5}$
Multiply the whole equation by $5$ to avoid working with fractions:
$5\Big(\dfrac{2}{5}y^{2}+\dfrac{1}{5}y=\dfrac{3}{5}\Big)$
$2y^{2}+y=3$
Take the $3$ to the left side of the equation:
$2y^{2}+y-3=0$
Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=2$, $b=1$ and $c=-3$
Substitute:
$y=\dfrac{-1\pm\sqrt{1^{2}-4(2)(-3)}}{2(2)}=\dfrac{-1\pm\sqrt{1+24}}{4}=\dfrac{-1\pm\sqrt{25}}{4}=...$
$...=\dfrac{-1\pm5}{4}$
Our two solutions are:
$y=\dfrac{-1+5}{4}=1$
$y=\dfrac{-1-5}{4}=-\dfrac{3}{2}$