# How do you find the roots, real and imaginary, of y=2x^2 + 13x + 6+4(x -1)^2  using the quadratic formula?

Dec 29, 2015

$x = \frac{- 5 + \sqrt{- 215}}{12}$ or $x = \frac{- 5 - \sqrt{- 215}}{12}$

#### Explanation:

First expand and simplify to ensure just one term for each power of x.
$y = 2 {x}^{2} + 13 x + 6 + 4 \left({x}^{2} - 2 x + 1\right)$
$y = 2 {x}^{2} + 13 x + 6 + 4 {x}^{2} - 8 x + 4$
$y = 6 {x}^{2} + 5 x + 10$
Then use $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 6 \cdot 10}}{2 \cdot 6}$
$x = \frac{- 5 \pm \sqrt{25 - 240}}{12}$
$x = \frac{- 5 \pm \sqrt{- 215}}{12}$
There are only imaginary roots to this equation, meaning that in graphical terms the graph never intersects the x axis.