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

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Answer 1 · 2017-06-19T17:10:29

See a solution process below:

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

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

$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - (4 \cdot 1 \cdot 12)}}{2 \cdot 1}$ $x = (-(-4) +- sqrt((-4)^2 - (4 * 1 * 12)))/(2 * 1)$

$x = \frac{4 \pm \sqrt{16 - 48}}{2}$ $x = (4 +- sqrt(16 - 48))/2$

$x = \frac{4}{2} \pm \frac{\sqrt{16 - 48}}{2}$ $x = 4/2 +- sqrt(16 - 48)/2$

$x = 2 \pm \frac{\sqrt{16 - 48}}{2}$ $x = 2 +- sqrt(16 - 48)/2$

$x = 2 \pm \frac{\sqrt{- 32}}{2}$ $x = 2 +- sqrt(-32)/2$

$x = 2 \pm \frac{\sqrt{16 \cdot - 2}}{2}$ $x = 2 +- sqrt(16 * -2)/2$

$x = 2 \pm \frac{\sqrt{16} \cdot \sqrt{- 2}}{2}$ $x = 2 +- (sqrt(16) * sqrt(-2))/2$

$x = 2 \pm \frac{4 \cdot \sqrt{- 2}}{2}$ $x = 2 +- (4 * sqrt(-2))/2$

$x = 2 \pm 2 \sqrt{- 2}$ $x = 2 +- 2sqrt(-2)$

Or

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

How do you solve x2−4x+12=0x^2-4x +12=0 ?