3b-(2-b)=(3/2)(b+12)

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Solution for 3b-(2-b)=(3/2)(b+12) equation: 



3b-(2-b)=(3/2)(b+12)

We move all terms to the left:

3b-(2-b)-((3/2)(b+12))=0



Domain of the equation: 2)(b+12))!=0

b∈R



We add all the numbers together, and all the variables

3b-(-1b+2)-((+3/2)(b+12))=0

We get rid of parentheses

3b+1b-((+3/2)(b+12))-2=0

We multiply parentheses ..

-((+3b^2+3/2*12))+3b+1b-2=0

We multiply all the terms by the denominator


-((+3b^2+3+3b*2*12))+1b*2*12))-2*2*12))=0



We calculate terms in parentheses: -((+3b^2+3+3b*2*12)), so:

(+3b^2+3+3b*2*12)

We get rid of parentheses

3b^2+3b*2*12+3

Wy multiply elements

3b^2+72b*1+3

Wy multiply elements

3b^2+72b+3

Back to the equation:

-(3b^2+72b+3)



Wy multiply elements

576b^3*1-(3b^2+72b+3)=0

We get rid of parentheses

576b^3*1-3b^2-72b-3=0

We do not support ebpression: b^3

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