# How do you find the roots, real and imaginary, of y= x^2 -3x +(-x-2)^2  using the quadratic formula?

Nov 5, 2017

$y$ has two complex roots: $x = \frac{1}{4} \left(- 1 \pm \sqrt{31} i\right)$

#### Explanation:

$y = {x}^{2} - 3 x + {\left(- x - 2\right)}^{2}$

The roots of $y$ occur where $y = 0$

$\therefore {x}^{2} - 3 x + {\left(- x - 2\right)}^{2} = 0$

Expanding:

${x}^{2} - 3 x + {x}^{2} + 4 x + 4 = 0$

Combining like terms:

$2 {x}^{2} + x + 4 = 0$

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

$= \frac{- 1 \pm \sqrt{1 - 32}}{4}$

$= \frac{- 1 \pm \sqrt{- 31}}{4}$

$= \frac{- 1 \pm \sqrt{31 \times \left(- 1\right)}}{4}$

$= \frac{1}{4} \left(- 1 \pm \sqrt{31} i\right)$

$y$ has two complex roots: $x = \frac{1}{4} \left(- 1 \pm \sqrt{31} i\right)$