Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 637: 74

{$-3,\dfrac{5}{2}$}

Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ Given: $2x^2+x=1 5$ It can be written as: $2x^2+x-15=0$ Thus, $x=\dfrac{-(1) \pm \sqrt{(1)^2-4(2)(-15)}}{2(2)}$ or, $x=\dfrac{-1 \pm \sqrt{121}}{4}$ or, $x=\dfrac{-1+11}{4}=\dfrac{5}{2}$ and $x=\dfrac{-1-11}{4}=-3$ Our solution set is: {$-3,\dfrac{5}{2}$}