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z^2=z+1
We move all terms to the left:
z^2-(z+1)=0
We get rid of parentheses
z^2-z-1=0
We add all the numbers together, and all the variables
z^2-1z-1=0
a = 1; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·1·(-1)
Δ = 5
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{5}}{2*1}=\frac{1-\sqrt{5}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{5}}{2*1}=\frac{1+\sqrt{5}}{2} $
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