54=-18t^2+72t

Solution for 54=-18t^2+72t equation:

54=-18t^2+72t

We move all terms to the left:

54-(-18t^2+72t)=0

We get rid of parentheses

18t^2-72t+54=0

a = 18; b = -72; c = +54;
Δ = b2-4ac
Δ = -722-4·18·54
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-36}{2*18}=\frac{36}{36}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+36}{2*18}=\frac{108}{36}
=3
$