Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 20

{$\dfrac{-2+\sqrt{7}}{5},\dfrac{-2-\sqrt{7}}{5}$}

Given: $(x+\dfrac{2}{5})^2=\dfrac{7}{25}$ Apply The Square Root Property. This can be written as: $(x+\dfrac{2}{5})=\sqrt{\dfrac{7}{25}}$ or, $(x+\dfrac{2}{5})=\pm \dfrac{\sqrt{7}}{5}$ or, $x=-\dfrac{2}{5}+\dfrac{\sqrt{7}}{5},-\dfrac{2}{5}-\dfrac{\sqrt{7}}{5}$ Hence, solution set is $x=${$\dfrac{-2+\sqrt{7}}{5},\dfrac{-2-\sqrt{7}}{5}$}