Solution
Gievn:
[tex]h=-2t^2-2t+12[/tex]When the rocket hits the ground. The distance is zero
Set h = 0 and solve for t
[tex]\begin{gathered} -2t^2-2t+12=0 \\ 2t^2+2t-12=0 \\ 2t^2+6t-4t-12=0 \\ 2t(t+3)-4(t+3)=0 \\ (t+3)(2t-4)=0 \end{gathered}[/tex][tex]\begin{gathered} t+3=0\text{ or 2t-4=0} \\ t=-3\text{ or 2t=4} \\ t=-3\text{ or t=}\frac{4}{2}=2 \end{gathered}[/tex]But, time can not be in negative, hence the answer t = 2