How to put $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ in canonical form We are given the equation $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$
We did an example of this in class but the equation had less terms.
I took a note in class that says : if there are linear terms, I have to rotate...
This is what I think I have to do. 


*

*Put the coefficients of this equation in a matrix $A$, then $B =
    det(A - I \lambda)$

*$det(B)$ must be $ = 0$

*Evaluating this determinant will give me a value for lambda that is
the root of the equation.

*I can then go on and find eigen lines or eigen vectors.


My questions : 


*

*Am I right about the steps to solve it ? Shouldn't it give me a canonical form at the end ? because it my notes, it doesn't.

*How do I place the coefficients in the matrix A ?
this is my try : 
$A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$
the first 2 is for the $2x^2$ term, the second 2 next are for the $4xy$ term which I divided by 2...( I dont really know why btw :\ ) and the 6 is for the $6y^2$ term. Now how do I place the 6x and 2y terms ?
$A = \begin{bmatrix}2 & 2\\2 & 6\\3 & 1 \\\end{bmatrix}$
I add another line with those 2 divided by 2 ?
edit :
Here's my current work
I rewrote A as :
$ \begin{bmatrix} x & y \\\end{bmatrix} $ 
$ A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$
$ \begin{bmatrix} x \\ y \\\end{bmatrix} $ +
$ \begin{bmatrix} 6 & 2 \\\end{bmatrix} $
$ \begin{bmatrix} x \\ y \\\end{bmatrix} $ - 6 
$B = \begin{bmatrix}2 - \lambda & 2\\2 & 6 - \lambda \\\end{bmatrix}$
$det(B) = (2 - \lambda)(6 - \lambda) - 4$
$det(B) = 12 - 2 \lambda - 6 \lambda + \lambda^2 - 4 $
And I'm stuck here. How do I factor this ?
$\lambda^2 - 8 \lambda + 8$
edit : so with the quadration formula I found two roots.
$4 + \sqrt{\frac{32}{2}}$ and 
$4 - \sqrt{\frac{32}{2}}$

I know of another method, but it may be equivalent to yours: you multiply
$A = \begin{bmatrix}cos\theta & sin\theta\\-sin\theta & cos\theta \\\end{bmatrix}$$  \begin{bmatrix}x\\ y \\\end{bmatrix}$ = $\begin{bmatrix}x' \\y' \\\end{bmatrix}$
And then sub in $(x',y')$ for $(x,y)$ in your original equation, set the $xy$ terms equal to $0$. The solution is the angle $\theta$ by which you must rotate the plane so that the $xy$-terms disappear. 
