H=-16t2+18t+5

Solution for H=-16t2+18t+5 equation:

=-16H^2+18H+5

We move all terms to the left:

-(-16H^2+18H+5)=0

We get rid of parentheses

16H^2-18H-5=0

a = 16; b = -18; c = -5;
Δ = b2-4ac
Δ = -182-4·16·(-5)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{161}}{2*16}=\frac{18-2\sqrt{161}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{161}}{2*16}=\frac{18+2\sqrt{161}}{32}
$