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5n^2+n-500=0
a = 5; b = 1; c = -500;
Δ = b2-4ac
Δ = 12-4·5·(-500)
Δ = 10001
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{-(1)-\sqrt{10001}}{2*5}=\frac{-1-\sqrt{10001}}{10} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{10001}}{2*5}=\frac{-1+\sqrt{10001}}{10} $
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