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n^2-19n-160=0
a = 1; b = -19; c = -160;
Δ = b2-4ac
Δ = -192-4·1·(-160)
Δ = 1001
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{-(-19)-\sqrt{1001}}{2*1}=\frac{19-\sqrt{1001}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1001}}{2*1}=\frac{19+\sqrt{1001}}{2} $
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