(x+7)(x-2)+(x+2)x-4)=(5+x)^2-53

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Solution for (x+7)(x-2)+(x+2)x-4)=(5+x)^2-53 equation: 



(x+7)(x-2)+(x+2)x-4)=(5+x)^2-53

We move all terms to the left:

(x+7)(x-2)+(x+2)x-4)-((5+x)^2-53)=0

We add all the numbers together, and all the variables

(x+7)(x-2)+(x+2)x-4)-((x+5)^2-53)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

x^2+(+x^2-2x+7x-14)+2x-4)-((x+5)^2=0

We get rid of parentheses

x^2+x^2-2x+7x+2x-4)-((x+5)^2-14=0

We add all the numbers together, and all the variables

2x^2+7x-4)-((x+5)^2-14=0

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

2x^2+7x-4)-((x+5)^2=14

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