H=-9t^2+27t

Solution for H=-9t^2+27t equation:

=-9H^2+27H

We move all terms to the left:

-(-9H^2+27H)=0

We get rid of parentheses

9H^2-27H=0

a = 9; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·9·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-27}{2*9}=\frac{0}{18}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*9}=\frac{54}{18}
=3
$