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z^2=256
We move all terms to the left:
z^2-(256)=0
a = 1; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·1·(-256)
Δ = 1024
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{1024}=32$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*1}=\frac{-32}{2} =-16 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*1}=\frac{32}{2} =16 $
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