Chapter 8 - Section 8.2 - The Quadratic Formula - Exercise Set - Page 607: 3

{$1,\dfrac{5}{2}$}

Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ Given: $2x^2-7x=-5$ It can be written as: $2x^2-7x+5=0$ Thus, $x=\dfrac{-(-7) \pm \sqrt{(-7)^2-4(2)(5)}}{2(2)}$ or, $x=\dfrac{7 \pm 3}{4}$ or, $x=\dfrac{7 + 3}{4}=\dfrac{5}{2}$ and $x=\dfrac{7 -3}{4}=1$ Our solution set is: {$1,\dfrac{5}{2}$}