H=8t^2+18t+5

Solution for H=8t^2+18t+5 equation:

=8H^2+18H+5

We move all terms to the left:

-(8H^2+18H+5)=0

We get rid of parentheses

-8H^2-18H-5=0

a = -8; b = -18; c = -5;
Δ = b2-4ac
Δ = -182-4·(-8)·(-5)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{41}}{2*-8}=\frac{18-2\sqrt{41}}{-16}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{41}}{2*-8}=\frac{18+2\sqrt{41}}{-16}
$