How do you solve x^2-4x +12=0 ?

Jun 19, 2017

See a solution process below:

Explanation:

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substituting $1$ for $a$; $- 4$ for $b$ and $12$ for $c$ gives:

$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \cdot 1 \cdot 12\right)}}{2 \cdot 1}$

$x = \frac{4 \pm \sqrt{16 - 48}}{2}$

$x = \frac{4}{2} \pm \frac{\sqrt{16 - 48}}{2}$

$x = 2 \pm \frac{\sqrt{16 - 48}}{2}$

$x = 2 \pm \frac{\sqrt{- 32}}{2}$

$x = 2 \pm \frac{\sqrt{16 \cdot - 2}}{2}$

$x = 2 \pm \frac{\sqrt{16} \cdot \sqrt{- 2}}{2}$

$x = 2 \pm \frac{4 \cdot \sqrt{- 2}}{2}$

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

Or

$2 \left(1 + \sqrt{- 2}\right)$