Answer
{$\frac{(-1 - i\sqrt {5})}{6},\frac{(-1 + i\sqrt {5})}{6}$}
Work Step by Step
Step 1: Comparing $6x^{2}+2x+1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=6$, $b=2$ 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:
$x=\frac{-(2) \pm \sqrt {(2)^{2}-4(6)(1)}}{2(6)}$
Step 4: $x=\frac{-2 \pm \sqrt {4-24}}{12}$
Step 5: $x=\frac{-2 \pm \sqrt {-20}}{12}$
Step 6: $x=\frac{-2 \pm \sqrt {-1\times20}}{12}$
Step 7: $x=\frac{-2 \pm (\sqrt {-1}\times\sqrt {4\times5})}{12}$
Step 8: $x=\frac{-2 \pm (i\times 2\sqrt {5})}{12}$
Step 9: $x=\frac{2(-1 \pm i\sqrt {5}}{12}$
Step 10: $x=\frac{(-1 \pm i\sqrt {5}}{6}$
Step 11: $x=\frac{(-1 - i\sqrt {5})}{6}$ or $x=\frac{(-1 + i\sqrt {5})}{6}$
Step 12: Therefore, the solution set is {$\frac{(-1 - i\sqrt {5})}{6},\frac{(-1 + i\sqrt {5})}{6}$}.
