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n(n+1)=840
We move all terms to the left:
n(n+1)-(840)=0
We multiply parentheses
n^2+n-840=0
a = 1; b = 1; c = -840;
Δ = b2-4ac
Δ = 12-4·1·(-840)
Δ = 3361
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{3361}}{2*1}=\frac{-1-\sqrt{3361}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3361}}{2*1}=\frac{-1+\sqrt{3361}}{2} $
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