## Algebra: A Combined Approach (4th Edition)

$y=1$ and $y=-\dfrac{3}{5}$
$2y=5y^{2}-3$ 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: $y=\dfrac{2+8}{10}=1$ $y=\dfrac{2-8}{10}=-\dfrac{3}{5}$