SOLUTION
We want to factor
[tex]\begin{gathered} 16x^2+8x=120 \\ 16x^2+8x-120=0 \end{gathered}[/tex]
We will solve this using the quadratic formula. We have
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ where \\ a=16,b=8,c=-120 \end{gathered}[/tex]
Applying we have
[tex]\begin{gathered} x=\frac{-8\pm\sqrt{8^2-4\times16\times-120}}{2\times16} \\ x=\frac{-8\pm\sqrt{64+7,680}}{32} \\ x=\frac{-8\pm\sqrt{7,744}}{32} \\ x=\frac{-8\pm88}{32} \end{gathered}[/tex]
Either
[tex]x=\frac{-8+88}{32}=\frac{80}{32}=2.5[/tex]
or
[tex]x=\frac{-8-88}{32}=\frac{-96}{32}=-3[/tex]
Hence the answer is x = 2.5, x = -3
