n(n-1)=800

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Solution for n(n-1)=800 equation:



n(n-1)=800
We move all terms to the left:
n(n-1)-(800)=0
We multiply parentheses
n^2-1n-800=0
a = 1; b = -1; c = -800;
Δ = b2-4ac
Δ = -12-4·1·(-800)
Δ = 3201
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{3201}}{2*1}=\frac{1-\sqrt{3201}}{2} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{3201}}{2*1}=\frac{1+\sqrt{3201}}{2} $

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