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

May 16, 2015

y = x^2 + 2x + 6.

This quadratic equation doesn't have real roots. It has 2 complex roots.

May 17, 2015

$0 = {x}^{2} + 2 x + 6 = {x}^{2} + 2 x + 1 + 5 = {\left(x + 1\right)}^{2} + 5$

Subtracting 5 from both sides we get

${\left(x + 1\right)}^{2} = - 5$

The square of any real number is positive or zero, so this has no solutions for real values of $x$.