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

Jan 28, 2017

$x = - \frac{8}{5}$

#### Explanation:

Let's expand and simplify the given equation:

$y = - {\left(x - 2\right)}^{2} + {x}^{2} + 11 x + 28$

$\textcolor{w h i t e}{y} = - \left({x}^{2} - 4 x + 4\right) + {x}^{2} + 11 x + 28$

$\textcolor{w h i t e}{y} = \textcolor{red}{\cancel{\textcolor{b l a c k}{- {x}^{2}}}} + 4 x - 4 + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + 11 x + 28$

$\textcolor{w h i t e}{y} = 15 x + 24$

Since the resulting equation is linear, not quadratic, the quadratic formula does not apply.

This linear equation has one zero, namely:

$x = - \frac{24}{15} = - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot 8}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot 5} = - \frac{8}{5}$