## Intermediate Algebra for College Students (7th Edition)

{$-\dfrac{2}{5},-\dfrac{4}{5}$}
Given: $x^2+\dfrac{6}{5}x+\dfrac{8}{25}=0$ This can be re-written as: $x^2+\dfrac{6}{5}x=-\dfrac{8}{25}$ We will have to add both sides $(\dfrac{3}{5})^2$ to complete the square. Thus, $x^2+\dfrac{6}{5}x+(\dfrac{3}{5})^2=-\dfrac{8}{25}+(\dfrac{3}{5})^2$ or, $(x+\dfrac{3}{5})^2=\dfrac{1}{25}$ or, $(x+\dfrac{3}{5})=\pm\sqrt{\dfrac{1}{25}}$ or, $x=-\dfrac{3}{5} \pm \dfrac{1}{5}$ Hence, our desired solution set is {$-\dfrac{2}{5},-\dfrac{4}{5}$}