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n2+n=2019
We move all terms to the left:
n2+n-(2019)=0
We add all the numbers together, and all the variables
n^2+n-2019=0
a = 1; b = 1; c = -2019;
Δ = b2-4ac
Δ = 12-4·1·(-2019)
Δ = 8077
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{8077}}{2*1}=\frac{-1-\sqrt{8077}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{8077}}{2*1}=\frac{-1+\sqrt{8077}}{2} $
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