2x+3x+3=x+(x+4)(+x+4)

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



2x+3x+3=x+(x+4)(+x+4)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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



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

x+(+x^2+4x+4x+16)

determiningTheFunctionDomain
(+x^2+4x+4x+16)+x

We get rid of parentheses

x^2+4x+4x+x+16

We add all the numbers together, and all the variables

x^2+9x+16

Back to the equation:

-(x^2+9x+16)



We add all the numbers together, and all the variables

5x-(x^2+9x+16)+3=0

We get rid of parentheses

-x^2+5x-9x-16+3=0

We add all the numbers together, and all the variables

-1x^2-4x-13=0

a = -1; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·(-1)·(-13)
Δ = -36
Delta is less than zero, so there is no solution for the equation

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