x(4x-3)+x(3x-3)=(7x+9)(x-2)

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Solution for x(4x-3)+x(3x-3)=(7x+9)(x-2) equation: 



x(4x-3)+x(3x-3)=(7x+9)(x-2)

We move all terms to the left:

x(4x-3)+x(3x-3)-((7x+9)(x-2))=0

We multiply parentheses

4x^2+3x^2-3x-3x-((7x+9)(x-2))=0

We multiply parentheses ..

4x^2+3x^2-((+7x^2-14x+9x-18))-3x-3x=0



We calculate terms in parentheses: -((+7x^2-14x+9x-18)), so:

(+7x^2-14x+9x-18)

We get rid of parentheses

7x^2-14x+9x-18

We add all the numbers together, and all the variables

7x^2-5x-18

Back to the equation:

-(7x^2-5x-18)



We add all the numbers together, and all the variables

7x^2-6x-(7x^2-5x-18)=0

We get rid of parentheses

7x^2-7x^2-6x+5x+18=0

We add all the numbers together, and all the variables

-1x+18=0

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

-x=-18

x=-18/-1

x=+18

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