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4n^2+9n-39=0
a = 4; b = 9; c = -39;
Δ = b2-4ac
Δ = 92-4·4·(-39)
Δ = 705
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{705}}{2*4}=\frac{-9-\sqrt{705}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{705}}{2*4}=\frac{-9+\sqrt{705}}{8} $
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