H(t)=18+6t-4t^2

Solution for H(t)=18+6t-4t^2 equation:

(H)=18+6H-4H^2

We move all terms to the left:

(H)-(18+6H-4H^2)=0

We get rid of parentheses

4H^2-6H+H-18=0

We add all the numbers together, and all the variables

4H^2-5H-18=0

a = 4; b = -5; c = -18;
Δ = b2-4ac
Δ = -52-4·4·(-18)
Δ = 313
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{-(-5)-\sqrt{313}}{2*4}=\frac{5-\sqrt{313}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{313}}{2*4}=\frac{5+\sqrt{313}}{8}
$