Question

16x^2+8x=120I need help please I forgot how to do factoring

Answer 1

We want to factor

`\begin{gathered} 16x^2+8x=120 \\ 16x^2+8x-120=0 \end{gathered}`

We will solve this using the quadratic formula. We have

`\begin{gathered} x=(-b\pm√(b^2-4ac))/(2a) \\ where \\ a=16,b=8,c=-120 \end{gathered}`

Applying we have

`\begin{gathered} x=(-8\pm√(8^2-4*16*-120))/(2*16) \\ x=(-8\pm√(64+7,680))/(32) \\ x=(-8\pm√(7,744))/(32) \\ x=(-8\pm88)/(32) \end{gathered}`

Either

`x=(-8+88)/(32)=(80)/(32)=2.5`

or

`x=(-8-88)/(32)=(-96)/(32)=-3`

Hence the answer is x = 2.5, x = -3