H=2+6t-2t^2

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

=2+6H-2H^2

We move all terms to the left:

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

We get rid of parentheses

2H^2-6H-2=0

a = 2; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·2·(-2)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*2}=\frac{6-2\sqrt{13}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*2}=\frac{6+2\sqrt{13}}{4}
$