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3n^2+3n-60=0
a = 3; b = 3; c = -60;
Δ = b2-4ac
Δ = 32-4·3·(-60)
Δ = 729
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{729}=27$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*3}=\frac{-30}{6} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*3}=\frac{24}{6} =4 $
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