# How do you solve 6x^2-12x+1=0 using the quadratic formula?

Aug 24, 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:

$\textcolor{red}{6}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 12}$ for $\textcolor{b l u e}{b}$

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

$x = \frac{- \textcolor{b l u e}{\left(- 12\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 12\right)}}^{2} - \left(4 \cdot \textcolor{red}{6} \cdot \textcolor{g r e e n}{1}\right)}}{2 \cdot \textcolor{red}{6}}$

$x = \frac{\textcolor{b l u e}{12} \pm \sqrt{144 - 24}}{12}$

$x = \frac{\textcolor{b l u e}{12} \pm \sqrt{120}}{12}$

$x = \frac{\textcolor{b l u e}{12} \pm \sqrt{4 \cdot 30}}{12}$

$x = \frac{\textcolor{b l u e}{12} \pm 2 \sqrt{30}}{12}$

Or

$x = 1 \pm \frac{\sqrt{30}}{6}$