H=-2t^2+20t+6

Solution for H=-2t^2+20t+6 equation:

=-2H^2+20H+6

We move all terms to the left:

-(-2H^2+20H+6)=0

We get rid of parentheses

2H^2-20H-6=0

a = 2; b = -20; c = -6;
Δ = b2-4ac
Δ = -202-4·2·(-6)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{7}}{2*2}=\frac{20-8\sqrt{7}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{7}}{2*2}=\frac{20+8\sqrt{7}}{4}
$