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n^2-3n-418=0
a = 1; b = -3; c = -418;
Δ = b2-4ac
Δ = -32-4·1·(-418)
Δ = 1681
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{1681}=41$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-41}{2*1}=\frac{-38}{2} =-19 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+41}{2*1}=\frac{44}{2} =22 $
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