# How do you solve x^2+8x-12=0?

Dec 12, 2016

$x = - 2 - 2 \sqrt{7}$ or $x = - 2 + 2 \sqrt{7}$

#### Explanation:

We can solve ${x}^{2} + 8 x - 12 = 0$ by using completing square method, as follows.

${x}^{2} + 8 x - 12 = 0$ can be written as

x^2+2×4×x+4^2-4^2-12=0

or ${\left(x + 2\right)}^{2} - 16 - 12 = 0$

or ${\left(x + 2\right)}^{2} - 28 = 0$

or ${\left(x + 2\right)}^{2} - {\left(\sqrt{28}\right)}^{2} = 0$

and as factors of ${a}^{2} - {b}^{2}$ are $\left(a + b\right)$ and $\left(a - b\right)$, quadratic polynomial on LHS can be factorized as under:

$\left(x + 2 + \sqrt{28}\right) \left(x + 2 - \sqrt{28}\right) = 0$

and hence either $x + 2 + \sqrt{28} = 0$ i.e. $x = - 2 - \sqrt{28} = - 2 - 2 \sqrt{7}$

or $x + 2 - \sqrt{28} = 0$ i.e. $x = - 2 + \sqrt{28} = - 2 + 2 \sqrt{7}$