# How do you solve x^2 - 4x = 5 using the quadratic formula?

Jun 1, 2017

color(blue)(x=-1 or x=5

#### Explanation:

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

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

${x}^{2} - 4 x = 5$

$\therefore {x}^{2} - 4 x - 5 = 0$

$\therefore a = 1 , b = - 4 , c = - 5$

$\therefore x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 5\right)}}{2 \left(1\right)}$

$\therefore x = \frac{4 \pm \sqrt{16 + 20}}{2}$

$\therefore x = \frac{4 \pm \sqrt{36}}{2}$

$\therefore x = \frac{4 \pm 6}{2}$

$\therefore x = \frac{4 + 6}{2}$

or:

$\therefore x = \frac{4 - 6}{2}$

$\therefore x = \frac{10}{2}$ or $x = \frac{- 2}{2}$

:.color(blue)(x=5 or color(blue)(x=-1

substitute color(blue)(x=-1

$\therefore {\left(- 1\right)}^{2} - 4 \left(- 1\right) = 5$

$\therefore 1 + 4 = 5$

:.color(blue)(color(blue)(5=5

substitute color(blue)(x=5

$\therefore {\left(5\right)}^{2} - 4 \left(5\right) = 5$

$\therefore 25 - 20 = 5$

:.color(blue)(5=5

Jun 1, 2017

$x = 5 , x = - 1$

#### Explanation:

Subtract $5$ from both sides.

${x}^{2} - 4 x - 5 = 0$

From this, $a = 1 , b = - 4 , c = - 5$

The quadratic formula is $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substituting the values

$\frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

$\frac{4 \pm \sqrt{16 + 20}}{2}$

$\frac{4 + \sqrt{36}}{2}$

$\frac{4}{2} + \frac{\sqrt{36}}{2} , \frac{4}{2} - \frac{\sqrt{36}}{2}$

$2 + \frac{\sqrt{36}}{2} , 2 - \frac{\sqrt{36}}{2}$

$2 + \frac{6}{2} , 2 - \frac{6}{2}$

$2 + 3 , 2 - 3$

$5 , - 1$