Answer
Solution set = $\displaystyle \{\frac{-1+\sqrt{41}}{10},\frac{-1-\sqrt{41}}{10}\}.$
Work Step by Step
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$5x^{2}+x-2=0 \quad\rightarrow \left\{\begin{array}{l}
a=5\\
b=1\\
c=-2
\end{array}\right.$
$x=\displaystyle \frac{-1\pm\sqrt{1^{2}-4(5)(-2)}}{2(5)}=\frac{-1\pm\sqrt{1+40}}{10}=\frac{-1\pm\sqrt{41}}{10}$
$x= \displaystyle \frac{-1+\sqrt{41}}{10}\qquad$ or$ \displaystyle \qquad x=\frac{-1-\sqrt{41}}{10}$
Solution set = $\displaystyle \{\frac{-1+\sqrt{41}}{10},\frac{-1-\sqrt{41}}{10}\}.$
