Find the type of the equation Given

$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0$

I need to find the type of the equation, for that i tried to get the canonical form
So, turn into lambdas
$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3 = 0$
from here i can't get the form which is looks like 
$\pm\xi^2 \pm \eta^2 \pm \zeta^2 =0$
any help or advice is much appreciated

The equation
$$\sum_{i,j=1}^na_{ij}u_{x_ix_j}+a(x,u,\nabla u)=0$$
can be reduced to canonical form by applying the non singular linear transformation  $\xi=B^Tx$, where B is a matrix such that the transformation $y=B\eta$ reduces the quadratic form $\sum_{i,j=1}^na_{ij}y_iy_j$ to canonical form.
Example. Reduce to canonical form
$$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0.$$
$$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3=
\frac{(4\lambda_1+3\lambda_2+5\lambda_3)^2}{4}-\frac{(\lambda_2+7\lambda_3)^2}{4}$$
Let
$$y_1=\frac{4\lambda_1+3\lambda_2+5\lambda_3}{2},\;y_2=\frac{\lambda_2+7\lambda_3}{2},\;y_3=\lambda_3$$
$\Rightarrow$
$$\lambda_1=\frac12y_1-\frac32y_2+4y_2,\;\lambda_2=2y_2-7y_2,\;\lambda_2=y_3$$
$\Rightarrow$
$$B^T=\begin{pmatrix}\frac{1}{2} & -\frac{3}{2} & 4\\
0 & 2 & -7\\
0 & 0 & 1\end{pmatrix},\quad B=\begin{pmatrix}\frac{1}{2} & 0 & 0\\
-\frac{3}{2} & 2 & 0\\
4 & -7 & 1\end{pmatrix}$$
Transform to canonical form is:
$$\xi=\frac{x}{2},\;\eta=-\frac{3}{2}x+2y,\;\zeta=4x-7 y+z$$
After reducing we get
$$u_{\xi\xi}-u_{\eta\eta}+2u=0$$
