# How do you solve (x-4)(x+3) = 6?

Feb 5, 2016

$x = \frac{1 \pm \sqrt{73}}{2}$

#### Explanation:

How to solve $\left(x - 4\right) \left(x + 3\right) = 6$

Step 1: Multiple the expand and combine like terms for the left side of the equation

$\left(x - 4\right) \left(x + 3\right) = 6$

$\left({x}^{2} - 4 x + 3 x - 12\right) = 6$

$\left({x}^{2} - x \pm 2\right) = 6$

Step 2-Set the equation equal to zero by subtracting both side of the equation by 6

${x}^{2} - x - 12 = 6$

$- 6 \text{ " } - 6$
$= = = = =$
color(red)(x^2 -x -18= 0

Step 3 Solve the equation using quadratic formula, since we can't factor it

Recall: Quadratic formula: $x = \frac{- b \pm \sqrt{{\left(b\right)}^{2} - 4 a c}}{2 a}$

color(red)(x^2 -x -18= 0

$a = 1 , b = - 1 , c = - 18$

Substitute into the quadratic formula, we get
x= (-(-1)+- sqrt((-1)^2-4(1)(-18)))/(2(1)

$x = \frac{1 \pm \sqrt{1 + 72}}{2}$

$x = \frac{1 \pm \sqrt{73}}{2}$