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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4H^2-1H-4=0

a = 4; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·4·(-4)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{65}}{2*4}=\frac{1-\sqrt{65}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*4}=\frac{1+\sqrt{65}}{8}
$