x^2+3x-4=(x+4)(x-1)

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



x^2+3x-4=(x+4)(x-1)

We move all terms to the left:

x^2+3x-4-((x+4)(x-1))=0

We multiply parentheses ..

x^2-((+x^2-1x+4x-4))+3x-4=0



We calculate terms in parentheses: -((+x^2-1x+4x-4)), so:

(+x^2-1x+4x-4)

We get rid of parentheses

x^2-1x+4x-4

We add all the numbers together, and all the variables

x^2+3x-4

Back to the equation:

-(x^2+3x-4)



We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2-x^2+3x-3x+4-4=0

We add all the numbers together, and all the variables

=0

x=0/1

x=0

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