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4n^2+3n-27=0
a = 4; b = 3; c = -27;
Δ = b2-4ac
Δ = 32-4·4·(-27)
Δ = 441
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{441}=21$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*4}=\frac{-24}{8} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*4}=\frac{18}{8} =2+1/4 $
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