252=n(n-1)

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Solution for 252=n(n-1) equation:



252=n(n-1)
We move all terms to the left:
252-(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+252=0
We add all the numbers together, and all the variables
-1n^2+n+252=0
a = -1; b = 1; c = +252;
Δ = b2-4ac
Δ = 12-4·(-1)·252
Δ = 1009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1009}}{2*-1}=\frac{-1-\sqrt{1009}}{-2} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1009}}{2*-1}=\frac{-1+\sqrt{1009}}{-2} $

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