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41n^2+39n=0
a = 41; b = 39; c = 0;
Δ = b2-4ac
Δ = 392-4·41·0
Δ = 1521
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{1521}=39$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-39}{2*41}=\frac{-78}{82} =-39/41 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+39}{2*41}=\frac{0}{82} =0 $
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