# How do you solve 6x^2−7x−3=0  using the quadratic formula?

May 24, 2018

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 $\textcolor{red}{a}$

$\textcolor{b l u e}{- 7}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 3}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 7} \pm \sqrt{{\textcolor{b l u e}{- 7}}^{2} - \left(4 \cdot \textcolor{red}{6} \cdot \textcolor{g r e e n}{- 3}\right)}}{2 \cdot \textcolor{red}{6}}$

$x = \frac{7 \pm \sqrt{49 - \left(- 72\right)}}{12}$

$x = \frac{7 \pm \sqrt{49 + 72}}{12}$

$x = \frac{7 \pm \sqrt{121}}{12}$

$x = \frac{7 \pm 11}{12}$

$x = \frac{7 - 11}{12}$; $x = \frac{7 + 11}{12}$

$x = \frac{- 4}{12}$; $x = \frac{18}{12}$

$x = - \frac{1}{3}$; $x = \frac{3}{2}$