H=-16t2+80t+18

Solution for H=-16t2+80t+18 equation:

=-16H^2+80H+18

We move all terms to the left:

-(-16H^2+80H+18)=0

We get rid of parentheses

16H^2-80H-18=0

a = 16; b = -80; c = -18;
Δ = b2-4ac
Δ = -802-4·16·(-18)
Δ = 7552
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:

$\sqrt{\Delta}=\sqrt{7552}=\sqrt{64*118}=\sqrt{64}*\sqrt{118}=8\sqrt{118}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-8\sqrt{118}}{2*16}=\frac{80-8\sqrt{118}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+8\sqrt{118}}{2*16}=\frac{80+8\sqrt{118}}{32}
$