Answer
{$\dfrac{-2 -\sqrt{14}}{2},\dfrac{-2 + \sqrt{14}}{2}$}
Work Step by Step
Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Since, $2x^2+4x-5=0$
Thus, $x=\dfrac{-(4) \pm \sqrt{(4)^2-4(2)(-5)}}{2(2)}$
or, $x=\dfrac{-4 \pm \sqrt{56}}{4}$
or, $x=\dfrac{-2 \pm \sqrt{14}}{2}$
or, $x=\dfrac{-2 -\sqrt{14}}{2},\dfrac{-2 + \sqrt{14}}{2}$
Our solution set is: {$\dfrac{-2 -\sqrt{14}}{2},\dfrac{-2 + \sqrt{14}}{2}$}
