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4n^2+11n-3=0
a = 4; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·4·(-3)
Δ = 169
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{169}=13$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*4}=\frac{-24}{8} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*4}=\frac{2}{8} =1/4 $
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