# How do you find the roots, real and imaginary, of y=-2(x +3)^2-7x+5 using the quadratic formula?

##### 1 Answer
Aug 30, 2017

x=−8.75780488547035, x=−0.7421951145296504

There are no imaginary solutions.

#### Explanation:

First, combine like terms and simplify the expression so it's in standard quadratic form:

$a {x}^{2} + b x + c$

The original expression:

$- 2 {\left(x + 3\right)}^{2} - 7 x + 5$

After FOILing $\left(x + 3\right)$:

$\left(x + 3\right) \left(x + 3\right)$

$= {x}^{2} + 6 x + 9$

After multiplying that by $- 2$:

$- 2 {x}^{2} - 12 x - 18$

After adding additional terms and simplifying:

$- 2 {x}^{2} - 19 x - 13$

To find the roots (AKA zeros), just plug it into the quadratic formula:

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

=(9\pm\sqrt{(-19)^2-4(-2)(-13)})/(2(-2)

$= 9 \setminus \pm \setminus \frac{\sqrt{257}}{- 4}$

x=−8.75780488547035, x=−0.7421951145296504

There are no imaginary solutions.