(x+1)^2-4^2=(x-3)(x+5)

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Solution for (x+1)^2-4^2=(x-3)(x+5) equation: 



(x+1)^2-4^2=(x-3)(x+5)

We move all terms to the left:

(x+1)^2-4^2-((x-3)(x+5))=0

We add all the numbers together, and all the variables

(x+1)^2-((x-3)(x+5))-16=0

We multiply parentheses ..

-((+x^2+5x-3x-15))+(x+1)^2-16=0



We calculate terms in parentheses: -((+x^2+5x-3x-15)), so:

(+x^2+5x-3x-15)

We get rid of parentheses

x^2+5x-3x-15

We add all the numbers together, and all the variables

x^2+2x-15

Back to the equation:

-(x^2+2x-15)



We get rid of parentheses

-x^2-2x+(x+1)^2+15-16=0

We add all the numbers together, and all the variables

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

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

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

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