# How do you solve x^2-2x+6=0 by completing the square?

May 10, 2015

Write it as:
${x}^{2} - 2 x = - 6$
Add and subtract $1$;
${x}^{2} - 2 x + 1 - 1 = - 6$
${x}^{2} - 2 x + 1 = - 6 + 1$
${\left(x - 1\right)}^{2} = - 5$
So:
$x - 1 = \pm \sqrt{- 5}$
$x = 1 \pm i \sqrt{5}$
These are two Complex Solutions where $i = \sqrt{- 1}$