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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-(5(x+5)-(-1x^2+4x-4x+16))-1x+(x+3)^2=0



We calculate terms in parentheses: -(5(x+5)-(-1x^2+4x-4x+16)), so:

5(x+5)-(-1x^2+4x-4x+16)

determiningTheFunctionDomain
-(-1x^2+4x-4x+16)+5(x+5)

We multiply parentheses

-(-1x^2+4x-4x+16)+5x+25

We get rid of parentheses

1x^2-4x+4x+5x-16+25

We add all the numbers together, and all the variables

x^2+5x+9

Back to the equation:

-(x^2+5x+9)



We add all the numbers together, and all the variables

-1x-(x^2+5x+9)+(x+3)^2=0

We get rid of parentheses

-x^2-1x-5x+(x+3)^2-9=0

We add all the numbers together, and all the variables

-1x^2-6x+(x+3)^2-9=0

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

-1x^2-6x+(x+3)^2=9

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