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n^2-n-3=0
We add all the numbers together, and all the variables
n^2-1n-3=0
a = 1; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·1·(-3)
Δ = 13
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{13}}{2*1}=\frac{1-\sqrt{13}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{13}}{2*1}=\frac{1+\sqrt{13}}{2} $
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