H=16t^2+18t+3

Solution for H=16t^2+18t+3 equation:

=16H^2+18H+3

We move all terms to the left:

-(16H^2+18H+3)=0

We get rid of parentheses

-16H^2-18H-3=0

a = -16; b = -18; c = -3;
Δ = b2-4ac
Δ = -182-4·(-16)·(-3)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{33}}{2*-16}=\frac{18-2\sqrt{33}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{33}}{2*-16}=\frac{18+2\sqrt{33}}{-32}
$