Answer
{$1- i\sqrt 2,1+i\sqrt 2$}
Work Step by Step
Step 1: Comparing $x^{2}-2x+3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=1$, $b=-2$ and $c=3$
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:
$x=\frac{-(-2) \pm \sqrt {(-2)^{2}-4(1)(3)}}{2(1)}$
Step 4: $x=\frac{2 \pm \sqrt {4-12}}{2}$
Step 5: $x=\frac{2 \pm \sqrt {-8}}{2}$
Step 6: $x=\frac{2 \pm \sqrt {-1\times4\times2}}{2}$
Step 7: $x=\frac{2 \pm (\sqrt {-1}\times\sqrt 4\times\sqrt 2)}{2}$
Step 8: $x=\frac{2 \pm (i\times 2\times\sqrt 2)}{2}$
Step 9: $x=\frac{2 \pm 2i\sqrt 2}{2}$
Step 10: $x=\frac{2(1 \pm i\sqrt 2)}{2}$
Step 11: $x=1 \pm i\sqrt 2$
Step 12: $x=1- i\sqrt 2$ or $x=1+ i\sqrt 2$
Step 13: Therefore, the solution set is {$1- i\sqrt 2,1+i\sqrt 2$}.
