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3n^2-31n+56=0
a = 3; b = -31; c = +56;
Δ = b2-4ac
Δ = -312-4·3·56
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-17}{2*3}=\frac{14}{6} =2+1/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+17}{2*3}=\frac{48}{6} =8 $
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