Answer
{$\frac{1 - i\sqrt {2}}{3},\frac{1 + i\sqrt {2}}{3}$}
Work Step by Step
Step 1: Comparing $3x^{2}-2x+1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=3$, $b=-2$ and $c=1$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(-2) \pm \sqrt {(-2)^{2}-4(3)(1)}}{2(3)}$
Step 4: $x=\frac{2 \pm \sqrt {4-12}}{6}$
Step 5: $x=\frac{2 \pm \sqrt {-8}}{6}$
Step 6: $x=\frac{2 \pm \sqrt {-1\times8}}{6}$
Step 7: $x=\frac{2 \pm (\sqrt {-1}\times\sqrt {4\times2})}{6}$
Step 8: $x=\frac{2 \pm (i\times 2\sqrt {2})}{6}$
Step 9: $x=\frac{1 \pm i\sqrt {2}}{3}$
Step 10: $x=\frac{1 - i\sqrt {2}}{3}$ or $x=\frac{1 + i\sqrt {2}}{3}$
Step 11: Therefore, the solution set is {$\frac{1 - i\sqrt {2}}{3},\frac{1 + i\sqrt {2}}{3}$}.
