Question

i can't figure out this problem. Factoring it doesn't work, and neither does the quadradic equation (it comes out with imaginary numbers). any ideas?
(3x+2)(x+4)=48

Answer 1

Hm. It's already factored, isn't it?

Answer 2

i mean solving for x.

Answer 3

Sorry for the downtime there -- the quadratic formula doesn't give imaginary results. I suspect that, much like I did initially, you're forgetting to include the sign on the `c' part of `4ac'.

Answer 4

(3x+2)(x+4)=48
3x^2+12x+2x+8=48
3x^2+14x-40=0
We look at the discriminant first because of the mention of complex solutions:
b^2-4ac=14^2-4*3*(-40)>0, so there should be two distinct (real) roots by quadratic formula.