# How do you find the roots, real and imaginary, of y= -x^2 - 12x + 4  using the quadratic formula?

Oct 2, 2017

See a solution process below:

#### Explanation:

To find the roots of the equation set $y$ equal to $0$ and solve for $x$.

$- {x}^{2} - 12 x + 4 = 0$

Now, we can use the quadratic equation to solve this problem:

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 $\textcolor{red}{a}$

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

$\textcolor{g r e e n}{4}$ for $\textcolor{g r e e n}{c}$ gives:

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

$x = \frac{12 \pm \sqrt{144 - \left(- 16\right)}}{- 2}$

$x = \frac{12}{-} 2 \pm \frac{\sqrt{144 + 16}}{- 2}$

$x = - 6 \pm \frac{\sqrt{160}}{- 2}$

$x = - 6 - \frac{\sqrt{16 \cdot 10}}{- 2}$ and $x = - 6 + \frac{\sqrt{16 \cdot 10}}{- 2}$

$x = - 6 - \frac{\sqrt{16} \sqrt{10}}{- 2}$ and $x = - 6 + \frac{\sqrt{16} \sqrt{10}}{- 2}$

$x = - 6 - \frac{4 \sqrt{10}}{- 2}$ and $x = - 6 + \frac{4 \sqrt{10}}{- 2}$

$x = - 6 - \left(- 2 \sqrt{10}\right)$ and $x = - 6 + \left(- 2 \sqrt{10}\right)$

$x = - 6 + 2 \sqrt{10}$ and $x = - 6 - 2 \sqrt{10}$