Chapter 8 - Section 8.2 - The Quadratic Formula - Exercise Set - Page 607: 17

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

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