H=5+18t+16t^2

Solution for H=5+18t+16t^2 equation:

=5+18H+16H^2

We move all terms to the left:

-(5+18H+16H^2)=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 = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*-16}=\frac{16}{-32}
=-1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*-16}=\frac{20}{-32}
=-5/8
$