# How do you solve x^2 – 5x – 1 = - 7 using the quadratic formula?

May 13, 2016

$x = \frac{5 \pm \sqrt{- 23}}{4}$

It has imaginary roots

#### Explanation:

Given -

${x}^{2} - 5 x - 1 = - 7$

Take the constant to the LHS.

$2 {x}^{2} - 5 x - 1 + 7 = 0$
$2 {x}^{2} - 5 x + 6 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - \left(4 \times 2 \times 6\right)}}{2 \times 2}$
$x = \frac{5 \pm \sqrt{25 - 48}}{4}$
$x = \frac{5 \pm \sqrt{- 23}}{4}$

It has imaginary roots