(5x+4)^2-(3x+5)(2x+1)=20x(x-2)+27

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Solution for (5x+4)^2-(3x+5)(2x+1)=20x(x-2)+27 equation: 



(5x+4)^2-(3x+5)(2x+1)=20x(x-2)+27

We move all terms to the left:

(5x+4)^2-(3x+5)(2x+1)-(20x(x-2)+27)=0

We multiply parentheses ..

-(+6x^2+3x+10x+5)+(5x+4)^2-(20x(x-2)+27)=0



We calculate terms in parentheses: -(20x(x-2)+27), so:

20x(x-2)+27

We multiply parentheses

20x^2-40x+27

Back to the equation:

-(20x^2-40x+27)



We get rid of parentheses

-6x^2-20x^2-3x-10x+(5x+4)^2+40x-5-27=0

We add all the numbers together, and all the variables

-26x^2+27x+(5x+4)^2-32=0

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

-26x^2+27x+(5x+4)^2=32

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