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n^2+29n-440=0
a = 1; b = 29; c = -440;
Δ = b2-4ac
Δ = 292-4·1·(-440)
Δ = 2601
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{2601}=51$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-51}{2*1}=\frac{-80}{2} =-40 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+51}{2*1}=\frac{22}{2} =11 $
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