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

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



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

We move all terms to the left:

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

We multiply parentheses ..

-((+x^2+3x+5x+15))+(x+4)^2-1=0



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

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

We get rid of parentheses

x^2+3x+5x+15

We add all the numbers together, and all the variables

x^2+8x+15

Back to the equation:

-(x^2+8x+15)



We get rid of parentheses

-x^2-8x+(x+4)^2-15-1=0

We add all the numbers together, and all the variables

-1x^2-8x+(x+4)^2-16=0

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

-1x^2-8x+(x+4)^2=16

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