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(n^2)+39n-348=0
a = 1; b = 39; c = -348;
Δ = b2-4ac
Δ = 392-4·1·(-348)
Δ = 2913
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{-(39)-\sqrt{2913}}{2*1}=\frac{-39-\sqrt{2913}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{2913}}{2*1}=\frac{-39+\sqrt{2913}}{2} $
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