# How do you solve x^2-4x-1=0 using the quadratic formula?

Jul 1, 2016

$x = \left(2 - \sqrt{5}\right)$ or $x = \left(2 + \sqrt{5}\right)$

#### Explanation:

$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
the quadratic formula tells us that
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \left(\textcolor{red}{a}\right)}$

The given equation: ${x}^{2} - 4 x - 1 = 0$ can be written explicitly in the general form as:
$\textcolor{w h i t e}{\text{XXX")x=color(red)(1)x^2+color(blue)(""(-4))+color(green)(} \left(- 1\right)} = 0$

So the solution is
color(white)("XXX")x=(-color(blue)(""(-4))+-sqrt(color(blue)(""(-4))^2-4color(red)(""(1))color(green)(""(-1))))/(2(color(red)(1))

$\textcolor{w h i t e}{\text{XXXX}} = \frac{4 \pm \sqrt{20}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = 2 \pm \sqrt{5}$