SOLUTION: Hey! Huge thanks to anyone that can help! Pretty Confused!! The height of a ball above the ground when the ball is thrown up from the bottom of a 64m building can be modelled by

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Question 906770: Hey! Huge thanks to anyone that can help! Pretty Confused!!
The height of a ball above the ground when the ball is thrown up from the bottom of a 64m building can be modelled by the equation h(t) = -4.9^2 + 35t + 1.2. How long is the ball visible to someone looking out of a fifth floor or higher window? Assume the fifth floor window is 15m above the ground.
Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
Plug in 15 for h(t)
There should be two times all the time between the ball would be visible
15 = -4.9^2 + 35t + 1.2.
0=-4.9^2 + 35t + -13.8

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=95452 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.418846215612299, 6.72401092724484. Here's your graph: