3x+6x(x+3)=3(2x^2+7x)

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Solution for 3x+6x(x+3)=3(2x^2+7x) equation: 



3x+6x(x+3)=3(2x^2+7x)

We move all terms to the left:

3x+6x(x+3)-(3(2x^2+7x))=0

We multiply parentheses

6x^2+3x+18x-(3(2x^2+7x))=0



We calculate terms in parentheses: -(3(2x^2+7x)), so:

3(2x^2+7x)

We multiply parentheses

6x^2+21x

Back to the equation:

-(6x^2+21x)



We add all the numbers together, and all the variables

6x^2+21x-(6x^2+21x)=0

We get rid of parentheses

6x^2-6x^2+21x-21x=0

We add all the numbers together, and all the variables

=0

x=0/1

x=0

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