Can someone help me with this math word problem please?

There it is again! Physics problems are being written by people who cannot distinguish between a rocket and a projectile. Notice that the vertical acceleration is constant and equal to gravitational acceleration. This would make sense only if the projectile were somehow given initial velocity of 288 feet per second, and the rocket failed to ignite.

The conditions are clear though. You want to know are what times the height is zero. The flight duration is the difference of those times. Solve this equation, and take the difference of the solutions:

h(t) = 0

There are two times at which the height is equal to 1152. They are the solutions to this equation:

h(t) = 1152

To find out when it will hit the ground you need to set the equation equal to zero.
For the next part set it equal to 1152

16t² - 288 t = 0
2t² - 36 t = 0
t² - 18 t = 0
t ( t - 18 ) = 0
t = 18 sec

1152 = - 16t² + 288 t
16t² - 288 t + 1152 = 0
2t² - 36 t + 144 = 0
t² - 18t + 72 = 0
[ t - 12 ] [ t - 6 ] = 0
t = 6 sec , t = 12 sec

Thank you everyone. :)

h(t) = -16t² + 288t

When the rocket hits the ground, h(t) = 0 :
0 = -16t² + 288t
16t² - 288t = 0
t² - 18t = 0
t (t - 18) = 0
t = 0 (rejected for the rocket has not yet shot upward when t = 0) or t = 18

When the rocket reaches a height of 1152 ft above the ground, h(t) = 1152 :
1152 = -16t² + 288t
16t² - 288t + 1152 = 0
t² - 18t + 72 = 0
(t - 6)(t - 12) = 0
t = 6 or t = 12

Ans:
It takes 18 seconds for the rocket to hit the ground.
It takes 6 seconds or 12 seconds when the height of the rocket is 1152 feet above the ground.
(When t = 6, the rocket is going up. When t = 12, the rocket is going down.)

1152 = -16t^2 + 288t
16t^2 - 288t + 1152 = 0
16 * (t^2 - 18t + 72) = 0
t^2 - 18t + 72 = 0
t^2 - 18t = -72
t^2 - 18t + 81 = 81 - 72
(t - 9)^2 = 9
t - 9 = -3 , 3
t = 9 - 3 , 9 + 3
t = 6 , 12

h(t) = 0
0 = -16t^2 + 288t
0 = 16t^2 - 288t
0 = t^2 - 18t
0 = t * (t - 18)
t = 0 , 18