## Algebra 1: Common Core (15th Edition)

$x={\dfrac{10}{3}}$ $x={\dfrac{-15}{18}}$
$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $18x^{2}-45x-50=0$.Therefore, $a=18$,$b=-45$ and $c=-50$: Substitute the values: $x=\dfrac{45±{\sqrt {-45^{2}-4(18)(-50)}}}{2(18)}$ =$\dfrac{45±{\sqrt {2025-(-3600)}}}{36}$ =$\dfrac{45± {\sqrt {5625}}}{36}$ =$\dfrac{45±75}{36}$ Separate the equation into plus and minus equations: $x=\dfrac{45+75}{36}={\dfrac{10}{3}}$ $x=\dfrac{45-75}{36}={\dfrac{-15}{18}}$