Answer
{$\frac{1 - 3i\sqrt {3}}{2},\frac{1 + 3i\sqrt {3}}{2}$}
Work Step by Step
Step 1: Comparing $x^{2}-x+7=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=-1$ and $c=7$
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:
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(1)(7)}}{2(1)}$
Step 4: $x=\frac{1 \pm \sqrt {1-28}}{2}$
Step 5: $x=\frac{1 \pm \sqrt {-27}}{2}$
Step 6: $x=\frac{1 \pm \sqrt {-1\times9\times3}}{2}$
Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {9}\times\sqrt 3)}{2}$
Step 8: $x=\frac{1 \pm (i\times 3\times\sqrt 3)}{2}$
Step 9: $x=\frac{1 \pm 3i\sqrt {3}}{2}$
Step 10: $x=\frac{1 - 3i\sqrt {3}}{2}$ or $x=\frac{1 + 3i\sqrt {3}}{2}$
Step 11: Therefore, the solution set is {$\frac{1 - 3i\sqrt {3}}{2},\frac{1 + 3i\sqrt {3}}{2}$}.
