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2z^2=2z
We move all terms to the left:
2z^2-(2z)=0
a = 2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·2·0
Δ = 4
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{4}=2$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*2}=\frac{0}{4} =0 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*2}=\frac{4}{4} =1 $
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