# How do you solve x^2 + 2x-7=0 by completing the square?

Apr 29, 2015

Move the constant term to the right side of the equation, then add whatever quantity is necessary to both sides so the left side is a square

${x}^{2} + 2 x - 7 = 0$

${x}^{2} + 2 x = 7$

If the first two terms of ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
are ${x}^{2} \mathmr{and} 2 x$
then $a = 1$

${x}^{2} + 2 x + 1 = 7 + 1$

${\left(x + 1\right)}^{2} = 8$

$x + 1 = \pm \sqrt{8}$

$x = - 1 \pm 2 \sqrt{2}$