How do you solve 3x^2 – x – 1 = 0 using the quadratic formula?

Aug 28, 2017

See a solution process below:

Explanation:

The quadratic formula states:

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:

$\textcolor{red}{3}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 1}$ for $\textcolor{b l u e}{b}$

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

$x = \frac{- \textcolor{b l u e}{\left(- 1\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 1\right)}}^{2} - \left(4 \cdot \textcolor{red}{3} \cdot \textcolor{g r e e n}{- 1}\right)}}{2 \cdot \textcolor{red}{3}}$

$x = \frac{1 \pm \sqrt{\textcolor{b l u e}{1 - \left(- 12\right)}}}{6}$

$x = \frac{1 \pm \sqrt{\textcolor{b l u e}{1} + 12}}{6}$

$x = \frac{1 \pm \sqrt{13}}{6}$