2(3x2x)3(x+4)=7x+18

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Solution for 2(3x2x)3(x+4)=7x+18 equation: 



2(3x^2x)3(x+4)=7x+18

We move all terms to the left:

2(3x^2x)3(x+4)-(7x+18)=0

We get rid of parentheses

23x^2x3(x+4)-7x-18=0

We move all terms containing x to the left, all other terms to the right

23x^2x3(x+4)-7x=18

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