Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 19

{$\dfrac{-3+\sqrt{11}}{4},\dfrac{-3-\sqrt{11}}{4}$}

Given: $(x+\dfrac{3}{4})^2=\dfrac{11}{16}$ Apply The Square Root Property. This can be written as: $(x+\dfrac{3}{4})=\sqrt{\dfrac{11}{16}}$ or, $(x+\dfrac{3}{4})=\pm \dfrac{\sqrt{11}}{4}$ or, $x=-\dfrac{3}{4}+\dfrac{\sqrt{11}}{4},-\dfrac{3}{4}-\dfrac{\sqrt{11}}{4}$ Hence, solution set is $x=${$\dfrac{-3+\sqrt{11}}{4},\dfrac{-3-\sqrt{11}}{4}$}