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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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



We calculate terms in parentheses: -((-1x^2-2x+2x+4)+3+x^2), so:

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

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2-1x^2-2x+2x+4+3

We add all the numbers together, and all the variables

7

Back to the equation:

-(7)



We add all the numbers together, and all the variables

4x-2=0

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

4x=2

x=2/4

x=1/2

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