X(3x-2)+x(3x-5)=(6x+7)(x-2)

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Solution for X(3x-2)+x(3x-5)=(6x+7)(x-2) equation: 



X(3X-2)+X(3X-5)=(6X+7)(X-2)

We move all terms to the left:

X(3X-2)+X(3X-5)-((6X+7)(X-2))=0

We multiply parentheses

3X^2+3X^2-2X-5X-((6X+7)(X-2))=0

We multiply parentheses ..

3X^2+3X^2-((+6X^2-12X+7X-14))-2X-5X=0



We calculate terms in parentheses: -((+6X^2-12X+7X-14)), so:

(+6X^2-12X+7X-14)

We get rid of parentheses

6X^2-12X+7X-14

We add all the numbers together, and all the variables

6X^2-5X-14

Back to the equation:

-(6X^2-5X-14)



We add all the numbers together, and all the variables

6X^2-7X-(6X^2-5X-14)=0

We get rid of parentheses

6X^2-6X^2-7X+5X+14=0

We add all the numbers together, and all the variables

-2X+14=0

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

-2X=-14

X=-14/-2

X=+7

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