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3n^2+7n-4=0
a = 3; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·3·(-4)
Δ = 97
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{-(7)-\sqrt{97}}{2*3}=\frac{-7-\sqrt{97}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{97}}{2*3}=\frac{-7+\sqrt{97}}{6} $
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