Answer
x=-$\frac{15}{18}$, $\frac{10}{3}$
Work Step by Step
The quadratic formula states that if a$x^{2}$+bx+c=0, and a$\ne$0, then
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If
18$x^{2}$-45x-50=0, then a=18, b=-45, and c=-50, and to solve for x, then all we must do is plug the numbers into the quadratic formula then simplify.
$x=\frac{-(-45)\pm\sqrt{(-45)^2-4(18)(-50)}}{2(18)}$
=$\frac{45\pm\sqrt{2025-(-3600)}}{36}$ (-45 squared=2025, 4*18*-50=-3600, and 2*18=36)
=$\frac{45\pm\sqrt{5625}}{36}$ (2025-(-3600)=5625)
=$\frac{45\pm75}{36}$ (The square root of 5625=75)
Since there is a plus minus symbol, we of course need to make sure that we compute both of our answers.
x=$\frac{45+75}{36}$=$\frac{120}{36}$=$\frac{10}{3}$
x=$\frac{45-75}{36}$=$\frac{-30}{36}$=-$\frac{15}{18}$
