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

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

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

We move all terms to the left:

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

We get rid of parentheses

2H^2-100H+H=0

We add all the numbers together, and all the variables

2H^2-99H=0

a = 2; b = -99; c = 0;
Δ = b2-4ac
Δ = -992-4·2·0
Δ = 9801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9801}=99$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-99)-99}{2*2}=\frac{0}{4}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-99)+99}{2*2}=\frac{198}{4}
=49+1/2
$