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

Aug 9, 2017

See a solution process below:

#### Explanation:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 4}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{1}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 4\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 4\right)}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{1}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{\textcolor{b l u e}{4} \pm \sqrt{\textcolor{b l u e}{16} - 4}}{2}$

$x = \frac{\textcolor{b l u e}{4} \pm \sqrt{12}}{2}$

$x = \frac{\textcolor{b l u e}{4} - \sqrt{4 \cdot 3}}{2}$ and $x = \frac{\textcolor{b l u e}{4} + \sqrt{4 \cdot 3}}{2}$

$x = \frac{\textcolor{b l u e}{4} - \sqrt{4} \sqrt{3}}{2}$ and $x = \frac{\textcolor{b l u e}{4} + \sqrt{4} \sqrt{3}}{2}$

$x = \frac{\textcolor{b l u e}{4} - 2 \sqrt{3}}{2}$ and $x = \frac{\textcolor{b l u e}{4} + 2 \sqrt{3}}{2}$

$x = \frac{\textcolor{b l u e}{4}}{2} - \frac{2 \sqrt{3}}{2}$ and $x = \frac{\textcolor{b l u e}{4}}{2} + \frac{2 \sqrt{3}}{2}$

$x = 2 - \sqrt{3}$ and $x = 2 + \sqrt{3}$