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2n^2+9n-500=0
a = 2; b = 9; c = -500;
Δ = b2-4ac
Δ = 92-4·2·(-500)
Δ = 4081
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{-(9)-\sqrt{4081}}{2*2}=\frac{-9-\sqrt{4081}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{4081}}{2*2}=\frac{-9+\sqrt{4081}}{4} $
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