Could someone please help me with this math problem

first for
A) u diff the equation
h = -5t^2 + 20t + 1.2
dh/dt = -10t + 20
peak means dh/dt = 0
therefore
0 = -10t +20
10t = 20
t = 2
when t = 2 is the peak
sub t = 2 into the equation
h = -5(2)^2 + 20*2 +1.2
h = -20 + 40 + 1.2
h = 21.2 meters
max height
u can solve a b c using this

then D)
ball hit the ground means the root of the equation
just factorize it
-b -+sqroot(b^2 -4ac )/2a
a = -5
b = 20
c =1.2

(-20 +- sqrt(20^2 - 4(-5)(1.2)) ) / 2(-5)(20)
=two answers

(-20 + sqrt(20^2 - 4(-5)(1.2)) ) / 2(-5)(20) = negative answer so drop it

(-20 - sqrt(20^2 - 4(-5)(1.2)) ) / 2(-5)(20) = positive answer
= 4.059

so at t = 4.059 its height = 0

h = -5t² + 20t +1.2
divide out the -5 ...... h /-5 = t² - 4t - .24
complete the square ........= t² - 4t + 4 - 4 - .24
Rewrite ...............................= (t - 2)² - 4.24
Remultiply ................... h = -5(t - 2)² + 21.2
There now you have it
a) maximum height is 21,2 meters
b) 1.2 meters
c) it took 2 seconds to go up, so it took 2 seconds to come down
....that's 4 seconds
d) -5(t - 2)² = -21.2 ...... (t - 2)² = 4.24 ........ t - 2 = 2.0591 .... t = 4.0591 sec