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1/3z-(3-5)=2(z+2)(z-5)
We move all terms to the left:
1/3z-(3-5)-(2(z+2)(z-5))=0
Domain of the equation: 3z!=0We add all the numbers together, and all the variables
z!=0/3
z!=0
z∈R
1/3z-(2(z+2)(z-5))-(-2)=0
We add all the numbers together, and all the variables
1/3z-(2(z+2)(z-5))+2=0
We multiply parentheses ..
-(2(+z^2-5z+2z-10))+1/3z+2=0
We multiply all the terms by the denominator
-((2(+z^2-5z+2z-10)))*3z+2*3z+1=0
Wy multiply elements
-((2(+z^2-5z+2z-10)))*3z+6z+1=0
We move all terms containing z to the left, all other terms to the right
-((2(+z^2-5z+2z-10)))*3z+6z=-1
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