H(t)=-2t^2+8t

Solution for H(t)=-2t^2+8t equation:

(H)=-2H^2+8H

We move all terms to the left:

(H)-(-2H^2+8H)=0

We get rid of parentheses

2H^2-8H+H=0

We add all the numbers together, and all the variables

2H^2-7H=0

a = 2; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2}=\frac{0}{4}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2}=\frac{14}{4}
=3+1/2
$