H=-3t^2+3t+18

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

=-3H^2+3H+18

We move all terms to the left:

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

We get rid of parentheses

3H^2-3H-18=0

a = 3; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·3·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*3}=\frac{-12}{6}
=-2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*3}=\frac{18}{6}
=3
$