(y+1)^2+(y-2)^2=y^2-y(3-y)+13

Solution for (y+1)^2+(y-2)^2=y^2-y(3-y)+13 equation:

(y+1)^2+(y-2)^2=y^2-y(3-y)+13

We move all terms to the left:

(y+1)^2+(y-2)^2-(y^2-y(3-y)+13)=0

We add all the numbers together, and all the variables

(y+1)^2+(y-2)^2-(y^2-y(-1y+3)+13)=0


We calculate terms in parentheses: -(y^2-y(-1y+3)+13), so:

y^2-y(-1y+3)+13

We multiply parentheses

y^2+1y^2-3y+13

We add all the numbers together, and all the variables

2y^2-3y+13

Back to the equation:

-(2y^2-3y+13)


We get rid of parentheses

-2y^2+(y+1)^2+(y-2)^2+3y-13=0

We add all the numbers together, and all the variables

-2y^2+3y+(y+1)^2+(y-2)^2-13=0

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

-2y^2+3y+(y+1)^2+(y-2)^2=13