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

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



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

We move all terms to the left:

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

We multiply parentheses ..

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



We calculate terms in parentheses: -((+x^2+2x-4x-8)), so:

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

We get rid of parentheses

x^2+2x-4x-8

We add all the numbers together, and all the variables

x^2-2x-8

Back to the equation:

-(x^2-2x-8)



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x+12=0

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

4x=-12

x=-12/4

x=-3

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