H(t)=-16t^2+288t

Solution for H(t)=-16t^2+288t equation:

(H)=-16H^2+288H

We move all terms to the left:

(H)-(-16H^2+288H)=0

We get rid of parentheses

16H^2-288H+H=0

We add all the numbers together, and all the variables

16H^2-287H=0

a = 16; b = -287; c = 0;
Δ = b2-4ac
Δ = -2872-4·16·0
Δ = 82369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{82369}=287$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-287)-287}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-287)+287}{2*16}=\frac{574}{32}
=17+15/16
$