H(t)=2t^2-4t

Solution for H(t)=2t^2-4t equation:

(H)=2H^2-4H

We move all terms to the left:

(H)-(2H^2-4H)=0

We get rid of parentheses

-2H^2+H+4H=0

We add all the numbers together, and all the variables

-2H^2+5H=0

a = -2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-2)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-2}=\frac{0}{-4}
=0
$