Pls math/quadratic equation

One
Complete the square.

h = - 16t^2 + 80t The best way to solve for t is to complete the square.
h = -16(t^2 - 5t)
h= - 16(t^2 - 5t + (5/2)^2) - (-16 * 25/4)
h= -16(t^2 - 5t + (5/2)^2 )^2 + 100
h = -16(t^2 - 5/2)^2 + 100

Here's what all that tells you.
1. It reached a maximum height of 100 feet.
2. It reaches a maximum height (the sign in front of the 16 is minus). That means that the quadratic is a minus.
3. The time needed to reach this maximum height is 5/2 seconds or 2.5 seconds.
4. Here's the biggy. The time it was in the air was 2 times as much as 2.5 second. The reason is that the ball has to come down again.

Summary
Max height = 100 feet. <<<< answer
Time in the air = 5 seconds <<<< answer

Just to confirm what I said, a graph is included.
You should see the following things.
The graph cuts the x axis at 5
The graph has a maximum of (2.5,100)
The graph also cuts the x axis at 0. This was the starting time of the ascent.
It's always a good idea to get a graph from Desmos if you can.

Sorry. I forgot the graph. It should be attached. 

The answer would he -4,-2 because the points are just on the opposite side. since the x axis is +4 it's going to he -4 and since the y axis is -2 it stays there because whaen rotated it stays in the third quadrant