H(t)=-16t^2+300t+200

Solution for H(t)=-16t^2+300t+200 equation:

(H)=-16H^2+300H+200

We move all terms to the left:

(H)-(-16H^2+300H+200)=0

We get rid of parentheses

16H^2-300H+H-200=0

We add all the numbers together, and all the variables

16H^2-299H-200=0

a = 16; b = -299; c = -200;
Δ = b2-4ac
Δ = -2992-4·16·(-200)
Δ = 102201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-299)-\sqrt{102201}}{2*16}=\frac{299-\sqrt{102201}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-299)+\sqrt{102201}}{2*16}=\frac{299+\sqrt{102201}}{32}
$