x(4x-5)=(4x+4)(x-3)

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Solution for x(4x-5)=(4x+4)(x-3) equation: 



x(4x-5)=(4x+4)(x-3)

We move all terms to the left:

x(4x-5)-((4x+4)(x-3))=0

We multiply parentheses

4x^2-5x-((4x+4)(x-3))=0

We multiply parentheses ..

4x^2-((+4x^2-12x+4x-12))-5x=0



We calculate terms in parentheses: -((+4x^2-12x+4x-12)), so:

(+4x^2-12x+4x-12)

We get rid of parentheses

4x^2-12x+4x-12

We add all the numbers together, and all the variables

4x^2-8x-12

Back to the equation:

-(4x^2-8x-12)



We add all the numbers together, and all the variables

4x^2-5x-(4x^2-8x-12)=0

We get rid of parentheses

4x^2-4x^2-5x+8x+12=0

We add all the numbers together, and all the variables

3x+12=0

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

3x=-12

x=-12/3

x=-4

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