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n^2+11n-500=0
a = 1; b = 11; c = -500;
Δ = b2-4ac
Δ = 112-4·1·(-500)
Δ = 2121
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{2121}}{2*1}=\frac{-11-\sqrt{2121}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{2121}}{2*1}=\frac{-11+\sqrt{2121}}{2} $
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