Answer
$x=1$ or $x=-\displaystyle \frac{5}{2}$
Work Step by Step
$ 2x^{2}+3x-5=0\qquad$... solve with the Quadractic formula. $a=2,\ b=3,\ c=-5$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $3,\ a$ for $2$ and $c$ for $-5$.
$ x=\displaystyle \frac{-3\pm\sqrt{(3)^{2}-4\cdot(-5)\cdot 2}}{2\cdot 2}\qquad$... simplify.
$x=\displaystyle \frac{-3\pm\sqrt{9+40}}{2}$
$x=\displaystyle \frac{-3\pm\sqrt{49}}{4}$
$ x=\displaystyle \frac{-3\pm 7}{4}\qquad$... the symbol $\pm$ indicates two solutions.
$x=\displaystyle \frac{-3+7}{4}$ or $x=\displaystyle \frac{-3-7}{4}$
$x=1$ or $x=-\displaystyle \frac{5}{2}$
