Answer
{$1 - 3i\sqrt 2,1 + 3i\sqrt 2$}
Work Step by Step
Step 1: Comparing $y^{2}-2y+19=0$ to the standard form of a quadratic equation, $ay^{2}+by+c=0$, we find:
$a=1$, $b=-2$ and $c=19$
Step 2: The quadratic formula is:
$y=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$y=\frac{-(-2) \pm \sqrt {(-2)^{2}-4(1)(19)}}{2(1)}$
Step 4: $y=\frac{2 \pm \sqrt {4-76}}{2}$
Step 5: $y=\frac{2 \pm \sqrt {-72}}{2}$
Step 6: $y=\frac{2 \pm \sqrt {-1\times72}}{2}$
Step 7: $y=\frac{2 \pm (\sqrt {-1}\times\sqrt {72})}{2}$
Step 8: $y=\frac{2 \pm (\sqrt {-1}\times\sqrt {36\times2})}{2}$
Step 9: $y=\frac{2 \pm (i\times 6\sqrt 2)}{2}$
Step 10: $y=\frac{2 \pm 6i\sqrt 2}{2}$
Step 11: $y=\frac{2(1 \pm 3i\sqrt 2)}{2}$
Step 12: $y=1 \pm 3i\sqrt 2$
Step 13: $y=1 - 3i\sqrt 2$ or $y=1 + 3i\sqrt 2$
Step 14: Therefore, the solution set is {$1 - 3i\sqrt 2,1 + 3i\sqrt 2$}.
