Answer
$\{-1,\frac{1}{3}\}$
Work Step by Step
Add $\frac{1}{3}$ to both sides.
$x^2+\dfrac{2}{3}x-\dfrac{1}{3}+\dfrac{1}{3}=0+\dfrac{1}{3}$
Simplify.
$x^2+\dfrac{2}{3}x=\dfrac{1}{3}$
Complete the square by adding $\left(\dfrac{\frac{2}{3}}{2}\right)^2=\dfrac{1}{9}$ to both sides.
$x^2+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{1}{3}+\dfrac{1}{9}$
Simplify.
$x^2+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{3+1}{9}$
$x^2+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{4}{9}$
Factor the left side.
$\left(x+\dfrac{1}{3}\right)^2=\dfrac{4}{9}$
Take the square root of both sides:
$x+\dfrac{1}{3}=\pm \sqrt {\dfrac{4}{9}}$
$x+\dfrac{1}{3}=\pm \dfrac{2}{3}$
Split the expression to obtain:
$x+\dfrac{1}{3}=\dfrac{2}{3}\quad $ or $\quad x+\dfrac{1}{3}=-\dfrac{2}{3}$
Subtract $\frac{1}{3}$ from both sides of both equations.
$x+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{3}-\dfrac{1}{3}\quad $ or $\quad x+\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}-\dfrac{1}{3}$
Simplify.
$x=\dfrac{1}{3}\quad $ or $\quad x=-1$
Hence, the solution set is $\{-1,\frac{1}{3}\}$.
