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

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



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

We move all terms to the left:

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

We multiply parentheses ..

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



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

3x(x+2)

We multiply parentheses

3x^2+6x

Back to the equation:

-(3x^2+6x)



We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-13x+(x-2)^2-4=0

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

-1x^2-13x+(x-2)^2=4

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