132=n(n-1)

Solution for 132=n(n-1) equation:

132=n(n-1)

We move all terms to the left:

132-(n(n-1))=0


We calculate terms in parentheses: -(n(n-1)), so:

n(n-1)

We multiply parentheses

n^2-1n

Back to the equation:

-(n^2-1n)


We get rid of parentheses

-n^2+1n+132=0

We add all the numbers together, and all the variables

-1n^2+n+132=0

a = -1; b = 1; c = +132;
Δ = b2-4ac
Δ = 12-4·(-1)·132
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-23}{2*-1}=\frac{-24}{-2}
=+12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+23}{2*-1}=\frac{22}{-2}
=-11
$