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
Hm. It's already factored, isn't it?
i mean solving for x.
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'.
(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.
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