A ball is thrown upward with an initial speed of 
24.5 m/s. When is it 19.6 m high? (two answers)

The formula is 

                   s = sO + vOt + at2/2

Since its initial speed is 24.5 m/s upward, sO = 24.5

Since we assume it starts on the ground, sO = 0

Since we want to know when its distance is 19.6 m, 
s = 19.6

Since we know that the acceleration is due to 
gravity, and that the ball is thrown up on Earth,
(and not on the moon or Mars), then a = g = -9.8 m/s2. 
So a = -9.8.

Plug these into

                   s = sO + vOt + at2/2

                19.6 = 0 + 24.5t + (-9.8)t2/2 

                19.6 = 24.5t - 4.9t2

4.9t2 - 24.5t + 19.6 = 0

Clear of decimals by multiplying through by 10,
that is, by moving the decimals one place to the
right in all the terms.

  49t2 - 245t + 196 = 0

Divide all terms by 49

        t2 - 5t + 4 = 0

Factor the trinomial on the left sides:

     (t - 4)(t - 5) = 0

Setting the first factor = 0

              t - 4 = 0
            
                  t = 4 seconds

Setting the second factor = 0

              t - 5 = 0
            
                  t = 5 seconds

So it reaches the height of 19.6 m after 4 seconds
(on its way up) and again after 5 seconds (on its
way down).

Edwin