Ahh...Owinner. the sign is greater and EQUAL to, so -5 and -1/2 are actually valid solutions there.
the answer is
\((-\infty, -5]\cup(-1, -\frac{1}{2}]\)
I'm not sure if you can use 1 in the first blank but here's what I got
\(3x^2+12x-4-2x^2+6x+7 = x^2+18x+3 =(x+9)^2-78 =1(x+9)^2+(-78)\)
