H=-16t^2+80t+200

Solution for H=-16t^2+80t+200 equation:

=-16H^2+80H+200

We move all terms to the left:

-(-16H^2+80H+200)=0

We get rid of parentheses

16H^2-80H-200=0

a = 16; b = -80; c = -200;
Δ = b2-4ac
Δ = -802-4·16·(-200)
Δ = 19200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-80\sqrt{3}}{2*16}=\frac{80-80\sqrt{3}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+80\sqrt{3}}{2*16}=\frac{80+80\sqrt{3}}{32}
$