Trying to find the zeros of a quadratic?

• Sep 16th 2009, 06:05 PM
nascar77
Trying to find the zeros of a 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, 06:18 PM
Prove It
Quote:

Originally Posted by nascar77

-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

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

Do you know about the discriminant?

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

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.

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

$\displaystyle = 361 - 192$

$\displaystyle = 169$.

Since this is positive, there are two solutions.

The solutions are given by

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

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

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

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

$\displaystyle = -\frac{3}{2}$ or $\displaystyle = -8$.
• Sep 16th 2009, 07:14 PM
nascar77
thank you