# How do you solve x^2-4x-5=0?

Mar 4, 2018

See a solution process below:

#### Explanation:

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

$\left(x + 1\right) \left(x - 5\right) = 0$

Now, equate each term on the left side of the equation to $0$ and solve for $x$:

Solution 1:

$x + 1 = 0$

$x + 1 - \textcolor{red}{1} = 0 - \textcolor{red}{1}$

$x + 0 = - 1$

$x = - 1$

Solution 2:

$x - 5 = 0$

$x - 5 + \textcolor{red}{5} = 0 + \textcolor{red}{5}$

$x - 0 = 5$

$x = 5$

The Solution Set Is:

$x = \left\{- 1 , 5\right\}$