(x+3)(x+3)-20=X2+5x+2

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Solution for (x+3)(x+3)-20=X2+5x+2 equation: 



(x+3)(x+3)-20=x2+5x+2

We move all terms to the left:

(x+3)(x+3)-20-(x2+5x+2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

-x^2+(+x^2+3x+3x+9)-5x-2-20=0

We add all the numbers together, and all the variables

-1x^2+(+x^2+3x+3x+9)-5x-22=0

We get rid of parentheses

-1x^2+x^2+3x+3x-5x+9-22=0

We add all the numbers together, and all the variables

x-13=0

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

x=13

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