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3n^2+31n-60=0
a = 3; b = 31; c = -60;
Δ = b2-4ac
Δ = 312-4·3·(-60)
Δ = 1681
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{1681}=41$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-41}{2*3}=\frac{-72}{6} =-12 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+41}{2*3}=\frac{10}{6} =1+2/3 $
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