(3+x)+(6+x)+(9+x)=(x+4)2+(x+1)2

Solution for (3+x)+(6+x)+(9+x)=(x+4)2+(x+1)2 equation:

(3+x)+(6+x)+(9+x)=(x+4)2+(x+1)2

We move all terms to the left:

(3+x)+(6+x)+(9+x)-((x+4)2+(x+1)2)=0

We add all the numbers together, and all the variables

(x+3)+(x+6)+(x+9)-((x+4)2+(x+1)2)=0

We get rid of parentheses

x+x+x-((x+4)2+(x+1)2)+3+6+9=0


We calculate terms in parentheses: -((x+4)2+(x+1)2), so:

(x+4)2+(x+1)2

We multiply parentheses

2x+2x+8+2

We add all the numbers together, and all the variables

4x+10

Back to the equation:

-(4x+10)


We add all the numbers together, and all the variables

3x-(4x+10)+18=0

We get rid of parentheses

3x-4x-10+18=0

We add all the numbers together, and all the variables

-1x+8=0

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

-x=-8

x=-8/-1

x=+8