(x-4)(6x^2+2x-4)+(3x+10)^2=(4x^2-x^2+5x+22)

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Solution for (x-4)(6x^2+2x-4)+(3x+10)^2=(4x^2-x^2+5x+22) equation: 



(x-4)(6x^2+2x-4)+(3x+10)^2=(4x^2-x^2+5x+22)

We move all terms to the left:

(x-4)(6x^2+2x-4)+(3x+10)^2-((4x^2-x^2+5x+22))=0



We calculate terms in parentheses: -((4x^2-x^2+5x+22)), so:

(4x^2-x^2+5x+22)

We get rid of parentheses

4x^2-x^2+5x+22

We add all the numbers together, and all the variables

3x^2+5x+22

Back to the equation:

-(3x^2+5x+22)



We get rid of parentheses

-3x^2-5x+(x-4)(6x^2+2x-4)+(3x+10)^2-22=0

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

-3x^2-5x+(x-4)(6x^2+2x-4)+(3x+10)^2=22

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