(2x+3)^2+(2x-3)=(8x+6)(x-1)+22

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Solution for (2x+3)^2+(2x-3)=(8x+6)(x-1)+22 equation: 



(2x+3)^2+(2x-3)=(8x+6)(x-1)+22

We move all terms to the left:

(2x+3)^2+(2x-3)-((8x+6)(x-1)+22)=0

We get rid of parentheses

(2x+3)^2+2x-((8x+6)(x-1)+22)-3=0

We multiply parentheses ..

-((+8x^2-8x+6x-6)+22)+(2x+3)^2+2x-3=0



We calculate terms in parentheses: -((+8x^2-8x+6x-6)+22), so:

(+8x^2-8x+6x-6)+22

We get rid of parentheses

8x^2-8x+6x-6+22

We add all the numbers together, and all the variables

8x^2-2x+16

Back to the equation:

-(8x^2-2x+16)



We add all the numbers together, and all the variables

2x-(8x^2-2x+16)+(2x+3)^2-3=0

We get rid of parentheses

-8x^2+2x+2x+(2x+3)^2-16-3=0

We add all the numbers together, and all the variables

-8x^2+4x+(2x+3)^2-19=0

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

-8x^2+4x+(2x+3)^2=19

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