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3n^2-108=0
a = 3; b = 0; c = -108;
Δ = b2-4ac
Δ = 02-4·3·(-108)
Δ = 1296
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{1296}=36$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*3}=\frac{-36}{6} =-6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*3}=\frac{36}{6} =6 $
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