(6x-4)(2x-1)=4x(3x-4)+20

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Solution for (6x-4)(2x-1)=4x(3x-4)+20 equation: 



(6x-4)(2x-1)=4x(3x-4)+20

We move all terms to the left:

(6x-4)(2x-1)-(4x(3x-4)+20)=0

We multiply parentheses ..

(+12x^2-6x-8x+4)-(4x(3x-4)+20)=0



We calculate terms in parentheses: -(4x(3x-4)+20), so:

4x(3x-4)+20

We multiply parentheses

12x^2-16x+20

Back to the equation:

-(12x^2-16x+20)



We get rid of parentheses

12x^2-12x^2-6x-8x+16x+4-20=0

We add all the numbers together, and all the variables

2x-16=0

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

2x=16

x=16/2

x=8

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