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

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



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

We move all terms to the left:

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



We calculate terms in parentheses: -(2(x^2-3x)+8), so:

2(x^2-3x)+8

We multiply parentheses

2x^2-6x+8

Back to the equation:

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



We get rid of parentheses

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

We add all the numbers together, and all the variables

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

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

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

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