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

Mar 9, 2016

The solutions are:

$x = 3$

$x = - 4$

#### Explanation:

${x}^{2} + x - 12 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:

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

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

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

$= 1 + 48 = 49$

The solutions are found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 1\right) \pm \sqrt{49}}{2 \cdot 1} = \frac{- 1 \pm 7}{2}$

$x = \frac{- 1 + 7}{2} = \frac{6}{2} = 3$

$x = \frac{- 1 - 7}{2} = - \frac{8}{2} = - 4$

The solutions are:

$x = 3$

$x = - 4$

Mar 9, 2016

$x = - 4 , 3$

#### Explanation:

${x}^{2} + x - 12 = 0$

This is in the form of a Quadratic equation
(in form $a {x}^{2} + b x + c = 0$)

Where

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

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

Substitute the values

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

$\rightarrow x = \frac{- 1 \pm \sqrt{1 - \left(- 48\right)}}{2}$

$\rightarrow x = \frac{- 1 \pm \sqrt{1 + 48}}{2}$

$\rightarrow x = \frac{- 1 \pm \sqrt{49}}{2}$

$\rightarrow x = \frac{- 1 \pm 7}{2}$

Now we have two values for $2$

$x = \frac{- 1 - 7}{2} , x = \frac{- 1 + 7}{2}$

If we solve for it,We get

$\Rightarrow x = - 4 , 3$