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