H(t)= -80t2+340t-260 (in feet)
How many seconds will pass before the ball hits the ground?
First of all, what kind of motion is this?
If t = 0, that is, initially, the ball was 260 below the ground.
Secondly, I don't recognize those units, definitely not on this earth.
anyway ...
-80t^2 + 340t - 260 = 0
divide each term by -20
4t^2 - 17t + 13 = 0
(4t - 13)(t - 1) = 0
t = 13/4 or t = 1
So your equation has solutions t = 1 or t = 4/3
I will leave it up to you to make sense of those two answers.