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9n^2+n-13=0
a = 9; b = 1; c = -13;
Δ = b2-4ac
Δ = 12-4·9·(-13)
Δ = 469
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{-(1)-\sqrt{469}}{2*9}=\frac{-1-\sqrt{469}}{18} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{469}}{2*9}=\frac{-1+\sqrt{469}}{18} $
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