(2x+4)^2-(10x+20)=3x^2+5x+8

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Solution for (2x+4)^2-(10x+20)=3x^2+5x+8 equation: 



(2x+4)^2-(10x+20)=3x^2+5x+8

We move all terms to the left:

(2x+4)^2-(10x+20)-(3x^2+5x+8)=0

We get rid of parentheses

-3x^2+(2x+4)^2-10x-5x-20-8=0

We add all the numbers together, and all the variables

-3x^2-15x+(2x+4)^2-28=0

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

-3x^2-15x+(2x+4)^2=28

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