4(z+3)-4=28(1/2z+1)

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Solution for 4(z+3)-4=28(1/2z+1) equation:



4(z+3)-4=28(1/2z+1)
We move all terms to the left:
4(z+3)-4-(28(1/2z+1))=0
Domain of the equation: 2z+1))!=0
z∈R
We multiply parentheses
4z-(28(1/2z+1))+12-4=0
We multiply all the terms by the denominator
4z*2z+12*2z+1))-4*2z+1))+1))-(28(1=0
Wy multiply elements
8z^2+24z+1))-4*2z+1))+1))-(28(1=0

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