(b+45)+b+3/2b+(2b=540

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Solution for (b+45)+b+3/2b+(2b=540 equation: 



(b+45)+b+3/2b+(2b=540

We move all terms to the left:

(b+45)+b+3/2b+(2b-(540)=0



Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R



We add all the numbers together, and all the variables

b+(b+45)+3/2b+(2b-540=0

We get rid of parentheses

b+b+3/2b+(2b-540+45=0

We multiply all the terms by the denominator


b*2b+b*2b+((2b-540+45)*2b+3=0

We add all the numbers together, and all the variables

b*2b+b*2b+((2b-495)*2b+3=0

Wy multiply elements

2b^2+2b^2+((2b-495)*2b+3=0

We add all the numbers together, and all the variables

4b^2+((2b-495)*2b+3=0

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