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3g^2-27=0
a = 3; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·3·(-27)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*3}=\frac{-18}{6} =-3 $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*3}=\frac{18}{6} =3 $
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