## Precalculus (6th Edition)

$\color{blue}{\left\{\dfrac{1}{3}- \dfrac{i\sqrt2}{3}, \dfrac{1}{3}+ \dfrac{i\sqrt2}{3}\right\}}$
Add $1$ to both sides of the equation: $$3x^2-2x+1=-1+1 \\3x^2-2x+1=0$$ The quadratic equation above has $a=3$, $b=-2$, and $c=1$. Solve the equation using the quadratic formula to obtain: $$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(3)(1)}}{2(3)} \\x=\dfrac{2\pm\sqrt{4-12}}{6} \\x=\dfrac{2\pm\sqrt{-8}}{6} \\x=\dfrac{2\pm\sqrt{4(-1)(2)}}{6} \\x=\dfrac{2\pm 2i\sqrt{2}}{6} \\x=\dfrac{2}{6}\pm \dfrac{2i\sqrt2}{6} \\x=\dfrac{1}{3}\pm \dfrac{i\sqrt2}{3}$$ Thus, the solution set is $\color{blue}{\left\{\dfrac{1}{3}- \dfrac{i\sqrt2}{3}, \dfrac{1}{3}+ \dfrac{i\sqrt2}{3}\right\}}$.