H(100)=-16t^2+100t

Solution for H(100)=-16t^2+100t equation:

(100)=-16H^2+100H

We move all terms to the left:

(100)-(-16H^2+100H)=0

We get rid of parentheses

16H^2-100H+100=0

a = 16; b = -100; c = +100;
Δ = b2-4ac
Δ = -1002-4·16·100
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-60}{2*16}=\frac{40}{32}
=1+1/4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+60}{2*16}=\frac{160}{32}
=5
$