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n2-n-121=0

We add all the numbers together, and all the variables

n^2-1n-121=0

a = 1; b = -1; c = -121;

Δ = b^{2}-4ac

Δ = -1^{2}-4·1·(-121)

Δ = 485

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{485}}{2*1}=\frac{1-\sqrt{485}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{485}}{2*1}=\frac{1+\sqrt{485}}{2} $

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