H=-2t^2+4t+6

Solution for H=-2t^2+4t+6 equation:

=-2H^2+4H+6

We move all terms to the left:

-(-2H^2+4H+6)=0

We get rid of parentheses

2H^2-4H-6=0

a = 2; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*2}=\frac{-4}{4}
=-1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*2}=\frac{12}{4}
=3
$