Answer
$y=1$ and $y=-\dfrac{3}{5}$
Work Step by Step
$2y=5y^{2}-3$
Take the $2y$ to the right side of the equation:
$5y^{2}-2y-3=0$
Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation $a=5$, $b=-2$ and $c=-3$
Substitute:
$y=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(5)(-3)}}{2(5)}=\dfrac{2\pm\sqrt{4+60}}{10}=...$
$...=\dfrac{2\pm\sqrt{64}}{10}=\dfrac{2\pm8}{10}$
Our two solutions are:
$y=\dfrac{2+8}{10}=1$
$y=\dfrac{2-8}{10}=-\dfrac{3}{5}$
