A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been lanuched.

My work (im lost)

H = 48t - 16t^2

Step 1 (t + 12) (t + 4t)

Step 2 cancel out t left with (-12) (4t)

Step 3 simplify -12/4 4t/4t

Step 4 i am so lost your help is appreciated proper steps please.

Thanks

Mr.H