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4z^2-81=0
a = 4; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·4·(-81)
Δ = 1296
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{1296}=36$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*4}=\frac{-36}{8} =-4+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*4}=\frac{36}{8} =4+1/2 $
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