# How do you find the zeros, real and imaginary, of y= x^2+12x+6  using the quadratic formula?

Aug 12, 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:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

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

$\textcolor{g r e e n}{6}$ for $\textcolor{g r e e n}{c}$ gives:

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

$x = \frac{- \textcolor{b l u e}{12} \pm \sqrt{144 - 24}}{2}$

$x = \frac{- \textcolor{b l u e}{12} \pm \sqrt{120}}{2}$

$x = \frac{- \textcolor{b l u e}{12} \pm \sqrt{4 \cdot 30}}{2}$

$x = \frac{- \textcolor{b l u e}{12} - \sqrt{4 \cdot 30}}{2}$ and $x = \frac{- \textcolor{b l u e}{12} + \sqrt{4 \cdot 30}}{2}$

$x = \frac{- \textcolor{b l u e}{12} - \sqrt{4} \sqrt{30}}{2}$ and $x = \frac{- \textcolor{b l u e}{12} + \sqrt{4} \sqrt{30}}{2}$

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

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

$x = - 6 - \sqrt{30}$ and $x = - 6 + \sqrt{30}$