x(3x+4)+1(2x-1)(3x+4)=3(2x-1)

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Solution for x(3x+4)+1(2x-1)(3x+4)=3(2x-1) equation: 



x(3x+4)+1(2x-1)(3x+4)=3(2x-1)

We move all terms to the left:

x(3x+4)+1(2x-1)(3x+4)-(3(2x-1))=0

We multiply parentheses

3x^2+4x+1(2x-1)(3x+4)-(3(2x-1))=0

We multiply parentheses ..

3x^2+1(+6x^2+8x-3x-4)+4x-(3(2x-1))=0



We calculate terms in parentheses: -(3(2x-1)), so:

3(2x-1)

We multiply parentheses

6x-3

Back to the equation:

-(6x-3)



We get rid of parentheses

3x^2+1(+6x^2+8x-3x-4)+4x-6x+3=0

We add all the numbers together, and all the variables

3x^2+1(+6x^2+8x-3x-4)-2x+3=0

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

3x^2+1(+6x^2+8x-3x-4)-2x=-3

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