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2n^2+n-480=0
a = 2; b = 1; c = -480;
Δ = b2-4ac
Δ = 12-4·2·(-480)
Δ = 3841
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{3841}}{2*2}=\frac{-1-\sqrt{3841}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3841}}{2*2}=\frac{-1+\sqrt{3841}}{4} $
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