H(t)=-16t^2+304t

Solution for H(t)=-16t^2+304t equation:

(H)=-16H^2+304H

We move all terms to the left:

(H)-(-16H^2+304H)=0

We get rid of parentheses

16H^2-304H+H=0

We add all the numbers together, and all the variables

16H^2-303H=0

a = 16; b = -303; c = 0;
Δ = b2-4ac
Δ = -3032-4·16·0
Δ = 91809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{91809}=303$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-303)-303}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-303)+303}{2*16}=\frac{606}{32}
=18+15/16
$