This can be written 4t²-16t-20=4(t²-4t-5)=0.
4(t-5)(t+1)=0, so t=5 seconds for the ball to hit the ground, because t-5=0, so t=5. (The other “solution” would be t+1=0, t=-1, which is one second before the ball was kicked! So we reject t=-1.)
(2) We have to solve h(t)=28, so:
16t²-32t-20=0=4(4t²-8t-5)=4(2t+1)(2t-5). To make this zero we use 2t-5=0, so 2t=5, t=5/2=2.5 (seconds). Just type 2.5 in the box.
To make this zero, 2t-3=0, 2t=3, t=1.5.
(4) h(t)=-16t²+64t-64=0, 16t²-64t+64=0=16(t²-4t+4)=16(t-2)².
So t-2=0, t=2 seconds for the rocket to fall to the ground.