(x^2-2x)^2-8=7(x^-2x)

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Solution for (x^2-2x)^2-8=7(x^-2x) equation: 



(x^2-2x)^2-8=7(x^-2x)

We move all terms to the left:

(x^2-2x)^2-8-(7(x^-2x))=0



We calculate terms in parentheses: -(7(x^-2x)), so:

7(x^-2x)

We multiply parentheses

7x-14x

We add all the numbers together, and all the variables

-7x

Back to the equation:

-(-7x)



We get rid of parentheses

(x^2-2x)^2+7x-8=0

We add all the numbers together, and all the variables

7x+(x^2-2x)^2-8=0

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

7x+(x^2-2x)^2=8

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