# How do you solve x^2 - 3x = 40?

Feb 16, 2016

$x = 8 , - 5$

#### Explanation:

color(blue)(x^2-3x=40

Subtract $40$ both sides

We get:

color(red)(x^2-3x-40=0

You can solve this both by Factoring and using the Quadratic formula:

1) First we can solve it by factoring:

color(red)(x^2-3x-40=0

Factor ${x}^{2} - 3 x - 40$

We get

$\rightarrow \left(x - 8\right) \left(x + 5\right) = 0$

If you solve it you get

rArrcolor(green)(x=8,-5

2) Quadratic formula

color(red)(x^2-3x-40=0

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

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

In this case $a = 1 , b = - 3 , c = - 40$

Substitute the values into the equation:

$\rightarrow x = \frac{- \left(- 3\right) \pm \sqrt{- {3}^{2} - 4 \left(1\right) \left(- 40\right)}}{2 \left(1\right)}$

$\rightarrow x = \frac{3 \pm \sqrt{9 - \left(- 160\right)}}{2}$

$\rightarrow x = \frac{3 \pm \sqrt{9 + 160}}{2}$

$\rightarrow x = \frac{3 \pm \sqrt{169}}{2}$

$\Rightarrow x = \frac{\textcolor{\in \mathrm{di} g o}{3 \pm 13}}{2}$

Now we have two solutions:

x=color(orange)((3+13)/2),color(violet)((3-13)/2

Solve for the first and then into the second:

rarrx=color(orange)((3+13)/2

rArrcolor(green)(x=16/2=8

For the second

rarrcolor(violet)(x=(3-13)/2

rArrcolor(green)(x=-10/2=-5