# How do you find the roots, real and imaginary, of y=x^2 + 11x +28  using the quadratic formula?

##### 1 Answer
Dec 13, 2015

$x = - 7 , - 4$

#### Explanation:

Quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

A quadratic equation is in the general form

$a {x}^{2} + b x + c$

So: $\left\{\begin{matrix}a = 1 \\ b = 11 \\ c = 28\end{matrix}\right.$

Plug these into the quadratic formula.

$x = \frac{- 11 \pm \sqrt{{11}^{2} - 4 \left(1\right) \left(28\right)}}{2 \left(1\right)}$

$x = \frac{- 11 \pm \sqrt{121 - 112}}{2}$

$x = \frac{- 11 \pm \sqrt{9}}{2}$

$x = \frac{- 11 \pm 3}{2}$

$\left\{\begin{matrix}x = \frac{- 11 + 3}{2} = \frac{- 8}{2} = \textcolor{b l u e}{- 4} \\ x = \frac{- 11 - 3}{2} = \frac{- 14}{2} = \textcolor{b l u e}{- 7}\end{matrix}\right.$