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3n^2+6n=0
a = 3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·3·0
Δ = 36
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{36}=6$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*3}=\frac{-12}{6} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*3}=\frac{0}{6} =0 $
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