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

Dec 21, 2015

$x = \frac{5 + \sqrt{29}}{2}$

$x = \frac{5 - \sqrt{29}}{2}$

#### Explanation:

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

Subtract $1$ from both sides.

${x}^{2} - 5 x - 1 = 0$ is a quadratic equation in standard form, $a {x}^{2} + b x + c$, where $a = 1 , b = - 5 , \mathmr{and} c = - 1$.

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

Substitute the values for $a , b , \mathmr{and} c$ into the formula.

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

Simplify.

$x = \frac{5 \pm \sqrt{25 - \left(- 4\right)}}{2}$

$x = \frac{5 \pm \sqrt{25 + 4}}{2}$

$x = \frac{5 \pm \sqrt{29}}{2}$

Solve for $x$

$x = \frac{5 + \sqrt{29}}{2}$

$x = \frac{5 - \sqrt{29}}{2}$