# How do you find the roots, real and imaginary, of y=-(2x+2)^2+x^2 + 9x-1  using the quadratic formula?

Sep 6, 2017

The roots are imaginary. $y = \frac{1 + i \sqrt{59}}{6}$ or $y = \frac{1 - i \sqrt{59}}{6}$

#### Explanation:

Firstly, expand

$y = - \left[4 {x}^{2} + 8 x + 4\right] + {x}^{2} + 9 x - 1$

$= - 4 {x}^{2} - 8 x - 4 + {x}^{2} + 9 x - 1$

Simplifying

$y = - 3 {x}^{2} + x - 5$

Using $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Here

$a = - 3$, $b = 1$, $c = - 5$

Then,

$y = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 3 \cdot 5}}{2 \cdot - 3}$

$= \frac{\cancel{-} 1 \pm \sqrt{1 - 60}}{\cancel{-}} 6$

$= \frac{1 \pm \sqrt{- 59}}{6}$

$= \frac{1 \pm \sqrt{59} i}{6}$

So the roots are:

$y = \frac{1 + i \sqrt{59}}{6}$ or $y = \frac{1 - i \sqrt{59}}{6}$

So the roots are imaginary.