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2n^2+n-465=0
a = 2; b = 1; c = -465;
Δ = b2-4ac
Δ = 12-4·2·(-465)
Δ = 3721
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3721}=61$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-61}{2*2}=\frac{-62}{4} =-15+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+61}{2*2}=\frac{60}{4} =15 $
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