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n(n-1)=45
We move all terms to the left:
n(n-1)-(45)=0
We multiply parentheses
n^2-1n-45=0
a = 1; b = -1; c = -45;
Δ = b2-4ac
Δ = -12-4·1·(-45)
Δ = 181
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{181}}{2*1}=\frac{1-\sqrt{181}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{181}}{2*1}=\frac{1+\sqrt{181}}{2} $
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