(x+4)^2=2x(5x-1)-7(x-2)

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Solution for (x+4)^2=2x(5x-1)-7(x-2) equation: 



(x+4)^2=2x(5x-1)-7(x-2)

We move all terms to the left:

(x+4)^2-(2x(5x-1)-7(x-2))=0



We calculate terms in parentheses: -(2x(5x-1)-7(x-2)), so:

2x(5x-1)-7(x-2)

We multiply parentheses

10x^2-2x-7x+14

We add all the numbers together, and all the variables

10x^2-9x+14

Back to the equation:

-(10x^2-9x+14)



We get rid of parentheses

-10x^2+(x+4)^2+9x-14=0

We add all the numbers together, and all the variables

-10x^2+9x+(x+4)^2-14=0

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

-10x^2+9x+(x+4)^2=14

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