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n^2+89n=0
a = 1; b = 89; c = 0;
Δ = b2-4ac
Δ = 892-4·1·0
Δ = 7921
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{7921}=89$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(89)-89}{2*1}=\frac{-178}{2} =-89 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(89)+89}{2*1}=\frac{0}{2} =0 $
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