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n^2+9n-480=0
a = 1; b = 9; c = -480;
Δ = b2-4ac
Δ = 92-4·1·(-480)
Δ = 2001
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{2001}}{2*1}=\frac{-9-\sqrt{2001}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{2001}}{2*1}=\frac{-9+\sqrt{2001}}{2} $
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