# Trying to find the zeros of a quadratic?

• Sep 16th 2009, 07:05 PM
nascar77
Trying to find the zeros of a quadratic?
I have this quadratic:

-2x^2-19x-24

I need to find its zeros and factor it but I havent had any luck. I also tried the quadratic formula but it produces negative roots? any help is welcomed
• Sep 16th 2009, 07:18 PM
Prove It
Quote:

Originally Posted by nascar77
I have this quadratic:

-2x^2-19x-24

I need to find its zeros and factor it but I havent had any luck. I also tried the quadratic formula but it produces negative roots? any help is welcomed

$-2x^2 - 19x - 24 = 0$.

Do you know about the discriminant?

$\Delta = b^2 - 4ac$.

If the discriminant is positive, we have two solutions, if it is 0 we have one solution, and if it is negative we don't have any solutions.

$\Delta = (-19)^2 - 4(-2)(-24)$

$= 361 - 192$

$= 169$.

Since this is positive, there are two solutions.

The solutions are given by

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$

$= \frac{19 \pm \sqrt{169}}{2(-2)}$

$= \frac{19 \pm 13}{-4}$

$= \frac{6}{-4}$ or $= \frac{32}{-4}$

$= -\frac{3}{2}$ or $= -8$.
• Sep 16th 2009, 08:14 PM
nascar77
thank you