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2/z+2/2z=2
We move all terms to the left:
2/z+2/2z-(2)=0
Domain of the equation: z!=0
z∈R
Domain of the equation: 2z!=0We calculate fractions
z!=0/2
z!=0
z∈R
4z/2z^2+2z/2z^2-2=0
We multiply all the terms by the denominator
4z+2z-2*2z^2=0
We add all the numbers together, and all the variables
6z-2*2z^2=0
Wy multiply elements
-4z^2+6z=0
a = -4; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-4)·0
Δ = 36
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{36}=6$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-4}=\frac{-12}{-8} =1+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-4}=\frac{0}{-8} =0 $
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