# How do you find the roots, real and imaginary, of y=(x +5 )^2+6x+3 using the quadratic formula?

Dec 7, 2015

$x 1 = - 14$
$x 2 = - 2$

#### Explanation:

$y = {\left(x + 5\right)}^{2} + 6 x + 3$
Expand square term: $y = {x}^{2} + 10 x + 25 + 6 x + 3$
Therefore $y = {x}^{2} + 16 x + 28$

This quadratic factorizes so there is no need of the Quadratic Formula:

$y = \left(x + 14\right) \left(x + 2\right)$ which has zeros at $x = - 14 \mathmr{and} - 2$

If you would like to use the Quadratic Formula:

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

In this case $a = 1 , b = 16 , c = 28$

Hence; $x = \frac{- 16 \pm \sqrt{256 - 112}}{2}$
$x = \frac{- 16 \pm \sqrt{144}}{2}$
$x = \frac{- 16 \pm 12}{2}$

$x = - 14 \mathmr{and} - 2$

Regarding your question on real or complex roots, the roots of a quadratic function with real coefficients will either both be real or both be complex. In this case the roots are real.