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3n^2+12n-15=0
a = 3; b = 12; c = -15;
Δ = b2-4ac
Δ = 122-4·3·(-15)
Δ = 324
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{324}=18$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18}{2*3}=\frac{-30}{6} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18}{2*3}=\frac{6}{6} =1 $
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