How do you solve $x^{2} - 3 x - 12 = 0$ $x^2-3x-12=0$ ?

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How do you solve $x^{2} - 3 x - 12 = 0$ $x^2-3x-12=0$ ?

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Answer 1 · 2017-04-24T20:58:45

$x \approx 5.275$ $x~~5.275$
$x \approx - 2.275$ $x~~-2.275$

Explanation:

There are many ways to solve a quadratic equation.
[Look here if you are lost on how to solve quadratic equations.](http://www.sosmath.com/algebra/solve/solve4/s43/s43.html)

When in doubt you can't go wrong with using the quadratic formula.

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

where $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

For $x^{2} - 3 x - 12 = 0$ $x^2-3x-12=0$

$a = 1$ $a=1$
$b = - 3$ $b=-3$
$c = - 12$ $c=-12$

Now just substitute into the quadratic formula

$x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (- 12)}}{2 (1)}$ $x=(-(-3)+-sqrt((-3)^2-4(1)(-12)))/(2(1))$

$x = \frac{3 \pm \sqrt{9 + 48}}{2}$ $x=(3+-sqrt(9+48))/(2)$

$x = \frac{3 \pm \sqrt{57}}{2}$ $x=(3+-sqrt(57))/(2)$

Therefore,

$x = \frac{3 + \sqrt{57}}{2}$ $x=(3+sqrt(57))/(2)$ $- - \to$ $--->$ $\approx 5.275$ $~~5.275$

OR

$x = \frac{3 - \sqrt{57}}{2}$ $x=(3-sqrt(57))/(2)$ $- - \to$ $--->$ $\approx - 2.275$ $~~-2.275$

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How do you solve $x^{2} - 3 x - 12 = 0$ $x^2-3x-12=0$ ?

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