(3x-4)(3x-4)-(x-7)(x-7)=8x(x-2)

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Solution for (3x-4)(3x-4)-(x-7)(x-7)=8x(x-2) equation: 



(3x-4)(3x-4)-(x-7)(x-7)=8x(x-2)

We move all terms to the left:

(3x-4)(3x-4)-(x-7)(x-7)-(8x(x-2))=0

We multiply parentheses ..

(+9x^2-12x-12x+16)-(x-7)(x-7)-(8x(x-2))=0



We calculate terms in parentheses: -(8x(x-2)), so:

8x(x-2)

We multiply parentheses

8x^2-16x

Back to the equation:

-(8x^2-16x)



We get rid of parentheses

9x^2-8x^2-12x-12x-(x-7)(x-7)+16x+16=0

We multiply parentheses ..

9x^2-8x^2-(+x^2-7x-7x+49)-12x-12x+16x+16=0

We add all the numbers together, and all the variables

x^2-(+x^2-7x-7x+49)-8x+16=0

We get rid of parentheses

x^2-x^2+7x+7x-8x-49+16=0

We add all the numbers together, and all the variables

6x-33=0

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

6x=33

x=33/6

x=5+1/2

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