54=n(n-3)

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

54=n(n-3)

We move all terms to the left:

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


We calculate terms in parentheses: -(n(n-3)), so:

n(n-3)

We multiply parentheses

n^2-3n

Back to the equation:

-(n^2-3n)


We get rid of parentheses

-n^2+3n+54=0

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = +54;
Δ = b2-4ac
Δ = 32-4·(-1)·54
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*-1}=\frac{-18}{-2}
=+9
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*-1}=\frac{12}{-2}
=-6
$