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+3n-60=0

a = 3; b = 3; c = -60;
Δ = b2-4ac
Δ = 32-4·3·(-60)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*3}=\frac{-30}{6}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*3}=\frac{24}{6}
=4
$