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n(2n+1)=6
We move all terms to the left:
n(2n+1)-(6)=0
We multiply parentheses
2n^2+n-6=0
a = 2; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·2·(-6)
Δ = 49
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*2}=\frac{-8}{4} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*2}=\frac{6}{4} =1+1/2 $
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