# How do you find the roots, real and imaginary, of y=-12x^2-2x -12 using the quadratic formula?

Dec 8, 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:

$\textcolor{red}{- 12}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 2}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 12}$ for $\textcolor{g r e e n}{c}$ gives:

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

$x = \frac{\textcolor{b l u e}{2} \pm \sqrt{\textcolor{b l u e}{4 - 576}}}{-} 24$

$x = - \frac{2 \pm \sqrt{- 572}}{24}$

$x = - \frac{2 \pm \sqrt{4 \cdot - 143}}{24}$

$x = - \frac{2 \pm \sqrt{4} \sqrt{- 143}}{24}$

$x = - \frac{2 \pm 2 \sqrt{- 143}}{24}$

$x = - \frac{1 \pm 1 \sqrt{- 143}}{12}$

$x = - \frac{1 \pm \sqrt{- 143}}{12}$