H=-16t^2+11t+5.5

Solution for H=-16t^2+11t+5.5 equation:

=-16H^2+11H+5.5

We move all terms to the left:

-(-16H^2+11H+5.5)=0

We get rid of parentheses

16H^2-11H-5.5=0

a = 16; b = -11; c = -5.5;
Δ = b2-4ac
Δ = -112-4·16·(-5.5)
Δ = 473
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{473}}{2*16}=\frac{11-\sqrt{473}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{473}}{2*16}=\frac{11+\sqrt{473}}{32}
$