# How do you find the roots, real and imaginary, of y= 2(x+5)^2+(x-4)^2  using the quadratic formula?

Sep 3, 2016

Zeros are $- 2 - 3 \sqrt{2} i$ and $- 2 + 3 \sqrt{2} i$

#### Explanation:

Quadratic formula gives the roots of $y = a {x}^{2} + b x + c$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

As $y = 2 {\left(x + 5\right)}^{2} + {\left(x - 4\right)}^{2}$

= $2 \left({x}^{2} + 10 x + 25\right) + {x}^{2} - 8 x + 16$

= $2 {x}^{2} + 20 x + 50 + {x}^{2} - 8 x + 16$

= $3 {x}^{2} + 12 x + 66$

Hence zeros are given by

x=(-12+-sqrt(12^2-4×3×66))/(2×3)

= $\frac{- 12 \pm \sqrt{144 - 792}}{6}$

= $\frac{- 12 \pm \sqrt{- 648}}{6}$

= $\frac{- 12 \pm 18 \sqrt{2} i}{6}$

= $- 2 \pm 3 \sqrt{2} i$

Hence, zeros are $- 2 - 3 \sqrt{2} i$ and $- 2 + 3 \sqrt{2} i$