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n^2+9n-194=0
a = 1; b = 9; c = -194;
Δ = b2-4ac
Δ = 92-4·1·(-194)
Δ = 857
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{-(9)-\sqrt{857}}{2*1}=\frac{-9-\sqrt{857}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{857}}{2*1}=\frac{-9+\sqrt{857}}{2} $
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