#### Answer

{$\dfrac{-3}{2} ,4$}

#### Work Step by Step

Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Since, $2x^2-5x-12=0$
Thus, $x=\dfrac{-(-5) \pm \sqrt{(-5)^2-4(2)(-12)}}{2(2)}$
or, $x=\dfrac{5 \pm \sqrt{121}}{4} \implies x= \dfrac{5 \pm 11}{4}$
or, $x=\dfrac{5 - 11}{4},\dfrac{5 + 11}{4}$
Our solution set is: {$\dfrac{-3}{2} ,4$}