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n^2+29n-552=0
a = 1; b = 29; c = -552;
Δ = b2-4ac
Δ = 292-4·1·(-552)
Δ = 3049
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{-(29)-\sqrt{3049}}{2*1}=\frac{-29-\sqrt{3049}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{3049}}{2*1}=\frac{-29+\sqrt{3049}}{2} $
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