Answer
{$\dfrac{-3 - \sqrt{89}}{2},\dfrac{-3 + \sqrt{89}}{2}$}
Work Step by Step
Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Given: $x^2+3x-20=0$
Thus, $x=\dfrac{-(3) \pm \sqrt{(3)^2-4(1)(-20)}}{2(1)}$
or, $x=\dfrac{-3 \pm \sqrt{89}}{2}$
or, $x=\dfrac{-3 - \sqrt{89}}{2}$ and $x=\dfrac{-3 + \sqrt{89}}{2}$
Our solution set is: {$\dfrac{-3 - \sqrt{89}}{2},\dfrac{-3 + \sqrt{89}}{2}$}
