Question 1169617
The height of the ball as a function of the time is represented by a parabola.
The vertex of the parabola with equation y = ax^2 + bx + c, is at x = -b/2a.
Thus, the time to reach maximum height = -20/-10 = 2 s. The ball is in the air for
twice that, or t = 4 s.  
The maximum height is given by H(2) = -5(2)^2 + 20(2) = -20 + 40 = 20 m
15 = -5t^2 + 20t -> -5(t^2 - 4t + 3) = 0. This factors as (t-3)(t-1) = 0.
Thus, there are two times when the ball is at 15 m, t = 1s and t = 3s.