Answer
{$2 - i\sqrt 3,2 + i\sqrt 3$}
Work Step by Step
Step 1: Comparing $x^{2}-4x+7=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=-4$ 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:
$x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(7)}}{2(1)}$
Step 4: $x=\frac{4 \pm \sqrt {16-28}}{2}$
Step 5: $x=\frac{4 \pm \sqrt {-12}}{2}$
Step 6: $x=\frac{4 \pm \sqrt {-1\times12}}{2}$
Step 7: $x=\frac{4 \pm (\sqrt {-1}\times\sqrt {12})}{2}$
Step 8: $x=\frac{4 \pm (\sqrt {-1}\times\sqrt {4\times3})}{2}$
Step 9: $x=\frac{4 \pm (i\times 2\sqrt 3)}{2}$
Step 10: $x=\frac{4 \pm i2\sqrt 3}{2}$
Step 11: $x=\frac{2(2 \pm i\sqrt 3)}{2}$
Step 12: $x=2 \pm i\sqrt 3$
Step 13: $x=2 - i\sqrt 3$ or $x=2 + i\sqrt 3$
Step 14: Therefore, the solution set is {$2 - i\sqrt 3,2 + i\sqrt 3$}.
