Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 49

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

Given: $2x^2+3x-5=0$ This can be re-written as: $x^2+\dfrac{3}{2}x=\dfrac{5}{2}$ We will have to add both sides $(\dfrac{3}{4})^2$ to complete the square. Thus, $x^2+\dfrac{3}{2}x+(\dfrac{3}{4})^2=\dfrac{5}{2}+(\dfrac{3}{4})^2$ or, $(x+\dfrac{3}{4})^2=\dfrac{49}{16}$ or, $(x+\dfrac{3}{4})=\pm\sqrt{\dfrac{49}{16}}$ or, $x=-\dfrac{3}{4} \pm \dfrac{7}{4}$ Hence, our desired solution set is {$-\dfrac{5}{2},1 $}