Answer
$\left\{\dfrac{-1-\sqrt {7}}{6},\dfrac{-1+\sqrt {7}}{6}\right\}$
Work Step by Step
Add $\frac{1}{2}$ to both sides.
$3x^2+x-\dfrac{1}{2}+\dfrac{1}{2}=0+\dfrac{1}{2}$
Simplify.
$3x^2+x=\dfrac{1}{2}$
Divide both sides by $3$.
$x^2+\dfrac{1}{3}x=\dfrac{1}{6}$
Complete the square by adding $\left(\dfrac{\frac{1}{3}}{2}\right)^2=\dfrac{1}{36}$ to both sides.
$x^2+\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{1}{6}+\dfrac{1}{36}$
Simplify.
$x^2+\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{6+1}{36}$
$x^2+\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{7}{36}$
Factor the left side.
$\left(x+\dfrac{1}{6}\right)^2=\dfrac{7}{36}$
Take the square root of both sdies.
$x+\dfrac{1}{6}=\pm \sqrt {\dfrac{7}{36}}$
$x+\dfrac{1}{6}=\pm\dfrac{\sqrt {7}}{6}$
Split the expression to obtain:
$x+\dfrac{1}{6}=\dfrac{\sqrt {7}}{6}\quad$ or $\quad x+\dfrac{1}{6}=-\dfrac{\sqrt {7}}{6}$
Subtract $\frac{1}{6}$ from both sides of both equations.
$x+\dfrac{1}{6}-\dfrac{1}{6}=\dfrac{\sqrt {7}}{6}-\dfrac{1}{6}\quad $ or $\quad x+\dfrac{1}{6}-\dfrac{1}{6}=-\dfrac{\sqrt {7}}{6}-\dfrac{1}{6}$
Simplify.
$x=\dfrac{-1\sqrt {7}}{6}\quad $ or $\quad x=\dfrac{-1-\sqrt {7}}{6}$
Hence, the solution set is $\left\{\frac{-1-\sqrt {7}}{6},\frac{-1+\sqrt {7}}{6}\right\}$.
