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6n^2+25n+14=0
a = 6; b = 25; c = +14;
Δ = b2-4ac
Δ = 252-4·6·14
Δ = 289
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{289}=17$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-17}{2*6}=\frac{-42}{12} =-3+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*6}=\frac{-8}{12} =-2/3 $
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