Answer
Solution set: $\{1+3i\sqrt 2, 1-3i \sqrt 2\}$
Work Step by Step
$x^{2}-2x+19=0$
Using quadratic formula,
$x= \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
Substituting $a=1, b=-2, c=19$
$x= \frac{-(-2)±\sqrt ((-2)^{2}-4(1)(19))}{2(1)}$
$x= \frac{2±\sqrt (4-76)}{2}$
$x= \frac{2±\sqrt -72}{2}$
Using $\sqrt -b = i\sqrt b$
$x= \frac{2±i\sqrt 72}{2}$
$x= \frac{2±i\sqrt 36 \times 2}{2}$
$x= \frac{2±6i\sqrt 2}{2}$
$x= \frac{2(1±3i\sqrt 2)}{2}$
$x=1±3i\sqrt 2$
$x=1+3i\sqrt 2$ or $x=1-3i\sqrt 2$
