Question

12t^2+17t=40

-40 -40

-40+12t^2+17t

(-8-3t) (5-4t)

-40+32t-15+12t^2

-40+17t+12t^2

i should have set each factor to zero..help

Answer 1

I don't know the question, so I will just guess the question is to find the value of t.

\[12t^2+17t=40\]\[12t^2+17t-40=0\]\[(3t+8)(4t-5)=0\]

\[3t+8=0\]or \[4t-5=0\]
\[3t=-8\]or \[4t=5\]
\[t=-8/3\]or \[t=5/4\]