H(t) = -5t^2+20t+1
what time does the ball reach the same height it was kicked at again? when does the ball reach its max height? what is the max height?

H(t) = -5t² + 20t + 1
-5t² + 20t + 1 = 0
t = -(20) +/- √((20)² - 4(-5)(1))
                      2(-5)
t = -20 +/- √(400 + 20)
                  -10
t = -20 +/- √(420)
              -10
t = -20 +/- 2√(105)
              -10
t = -20 + 2√(105)      t = -20 - 2√(105)
             -10                           -10
t = 2 - 0.2√(105)      t = 2 + 0.2√(105)
h(t) = -5t² + 20t + 1
h(2 - 0.2√(105)) = -5(2 - 0.2√(105))² + 20(2 - 0.2√(105)) + 1
h(2 - 0.2√(105)) = -5(2 - 0.2√(105))(2 - 0.2√(105)) + 20(2) - 20(0.2√(105)) + 1
h(2 - 0.2√(105)) = -5(4 - 0.4√(105) - 0.4√(105) + 0.04√(11025)) + 40 - 4√(105) + 1
h(2 - 0.2√(105)) = -5(4 - 0.8√(105) + 0.04(105)) + 40 + 1 - 4√(105)
h(2 - 0.2√(105)) = -5(4 - 0.8√(105) + 4.2) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(4 + 4.2 - 0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(8.2 - 0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(8.2) - 5(-0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -41 + 4√(105) + 41 - 4√(105)
h(2 - 0.2√(105)) = -41 + 41 + 4√(105) - 4√(105)
h(2 - 0.2√(105)) = 0 + 0
h(2 - 0.2√(105)) = 0
(t, h(t)) = (2 - 0.2√(105), 0)
or
h(t) = -5t² + 20t + 1
h(2 + 0.2√(105)) = -5(2 + 0.2√(105))² + 20(2 + 0.2√(105)) + 1
h(2 + 0.2√(105)) = -5(2 + 0.2√(105))(2 + 0.2√(105)) + 20(2) + 20(0.2√(105)) + 1
h(2 + 0.2√(105)) = -5(4 + 0.4√(105) + 0.4√(105) + 0.04√(11025)) + 40 + 4√(105) + 1
h(2 + 0.2√(105)) = -5(4 + 0.8√(105) + 0.04(105)) + 40 + 4√(105) + 1
h(2 + 0.2√(105)) = -5(4 + 0.8√(105) + 4.2) + 40 + 1 + 4√(105)
h(2 + 0.2√(105)) = -5(4 + 4.2 + 0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -5(8.2 + 0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -5(8.2) - 5(0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -41 - 4√(105) + 41 + 4√(105)
h(2 + 0.2√(105)) = -41 + 41 - 4√(105) + 4√(105)
h(2 + 0.2√(105)) = 0 + 0
h(2 + 0.2√(105)) = 0
(t, h(t)) = (2 + 0.2√(105), 0)

Cwill be your correct answer