(x+2)(x+4)=(x+1)(x+3)+15

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Solution for (x+2)(x+4)=(x+1)(x+3)+15 equation: 



(x+2)(x+4)=(x+1)(x+3)+15

We move all terms to the left:

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

We multiply parentheses ..

(+x^2+4x+2x+8)-((x+1)(x+3)+15)=0



We calculate terms in parentheses: -((x+1)(x+3)+15), so:

(x+1)(x+3)+15

We multiply parentheses ..

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

We get rid of parentheses

x^2+3x+x+3+15

We add all the numbers together, and all the variables

x^2+4x+18

Back to the equation:

-(x^2+4x+18)



We get rid of parentheses

x^2-x^2+4x+2x-4x+8-18=0

We add all the numbers together, and all the variables

2x-10=0

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

2x=10

x=10/2

x=5

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