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3z^2-243=0
a = 3; b = 0; c = -243;
Δ = b2-4ac
Δ = 02-4·3·(-243)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2916}=54$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-54}{2*3}=\frac{-54}{6} =-9 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+54}{2*3}=\frac{54}{6} =9 $
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