H=-10t^2+80t+92

Solution for H=-10t^2+80t+92 equation:

=-10H^2+80H+92

We move all terms to the left:

-(-10H^2+80H+92)=0

We get rid of parentheses

10H^2-80H-92=0

a = 10; b = -80; c = -92;
Δ = b2-4ac
Δ = -802-4·10·(-92)
Δ = 10080
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:

$\sqrt{\Delta}=\sqrt{10080}=\sqrt{144*70}=\sqrt{144}*\sqrt{70}=12\sqrt{70}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-12\sqrt{70}}{2*10}=\frac{80-12\sqrt{70}}{20}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+12\sqrt{70}}{2*10}=\frac{80+12\sqrt{70}}{20}
$