3(n-1)(n-2)=54

Solution for 3(n-1)(n-2)=54 equation:

3(n-1)(n-2)=54

We move all terms to the left:

3(n-1)(n-2)-(54)=0

We multiply parentheses ..

3(+n^2-2n-1n+2)-54=0

We multiply parentheses

3n^2-6n-3n+6-54=0

We add all the numbers together, and all the variables

3n^2-9n-48=0

a = 3; b = -9; c = -48;
Δ = b2-4ac
Δ = -92-4·3·(-48)
Δ = 657
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{73}}{2*3}=\frac{9-3\sqrt{73}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{73}}{2*3}=\frac{9+3\sqrt{73}}{6}
$