# How do you find the roots, real and imaginary, of y=x^2 + 14x +2(x/2-2)^2  using the quadratic formula?

Feb 17, 2016

$- 0.382$ and $- 2.618$

#### Explanation:

First distribute the 2 to the binomial.

$y = {x}^{2} + 14 x + 2 {\left(\frac{x}{2} - 2\right)}^{2}$

$y = {x}^{2} + 14 x + {\left(x - 4\right)}^{2}$

Next, square ${\left(x - 4\right)}^{2}$ using FOIL.

${x}^{2} + 14 x + {x}^{2} - 8 x + 16$

Combine like terms.

$2 {x}^{2} + 6 x + 16$

Divide all terms by $2$

${x}^{2} + 3 x + 8$

Now, using the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4}}{2} a$ where $a = 1 , b = 3 \mathmr{and} c = 8$

This gives $\frac{- 3 \pm \sqrt{{3}^{2} - 4}}{2 \cdot 1}$

==> $\frac{- 3 \pm \sqrt{9 - 4}}{2}$

==> $\frac{- 3 \pm \sqrt{5}}{2}$

This may be a good answer. If not, I'll use decimals...

==> $\frac{- 3 \pm 2.236}{2}$

==> $\frac{- 3 + 2.236}{2}$ = $- 0.382$

==> $\frac{- 3 - 2.236}{2}$ = $- 2.618$