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2n^2+13n+6=0

a = 2; b = 13; c = +6;

Δ = b^{2}-4ac

Δ = 13^{2}-4·2·6

Δ = 121

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{121}=11$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*2}=\frac{-24}{4} =-6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*2}=\frac{-2}{4} =-1/2 $

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