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3n^2+11n-500=0
a = 3; b = 11; c = -500;
Δ = b2-4ac
Δ = 112-4·3·(-500)
Δ = 6121
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{-(11)-\sqrt{6121}}{2*3}=\frac{-11-\sqrt{6121}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{6121}}{2*3}=\frac{-11+\sqrt{6121}}{6} $
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