H=-2t^2+6t+12

Solution for H=-2t^2+6t+12 equation:

=-2H^2+6H+12

We move all terms to the left:

-(-2H^2+6H+12)=0

We get rid of parentheses

2H^2-6H-12=0

a = 2; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·2·(-12)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*2}=\frac{6-2\sqrt{33}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*2}=\frac{6+2\sqrt{33}}{4}
$