# How do you factor and solve x^2-6x-5=0?

Jun 6, 2016

Use the quadratic formula to find

${x}^{2} - 6 x - 5 = \left(x - 3 - \sqrt{14}\right) \left(x - 3 + \sqrt{14}\right) = 0$

#### Explanation:

The first thing we notice is that $5$ is a prime number, so the only integer solutions which multiply to give $5$ would be $\pm 1$ and $\pm 5$, however, the only way to get these to add to be $- 6$ is if they are both negative, which results in $+ 5$, therefore, there are no integer solutions to factor.

The solution now is to use the quadratic formula to get the zeros of our equation and substitute them into the factored equation:

$\left(x - p\right) \left(x - q\right) = {x}^{2} - 6 x - 5 = 0$

Where we find $p$ and $q$ from

$p , q = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{6 \pm \sqrt{36 + 20}}{2} = 3 \pm \sqrt{14}$

so our equation becomes:

${x}^{2} - 6 x - 5 = \left(x - 3 - \sqrt{14}\right) \left(x - 3 + \sqrt{14}\right) = 0$