Answer
{$-\dfrac{2}{5},-\dfrac{4}{5}$}
Work Step by Step
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 {$-\dfrac{2}{5},-\dfrac{4}{5}$}
