Completing squares with three variables. I want to complete the squares for this polynomial 
$2x^2+2y^2-z^2+2xy+3xz-4yz$
Is there any kind of easy and non-confusing way to solve it?
I’ve done this up until now:
$$2x^2+2y^2-z^2+2xy+3xz-4yz$$
$$2x^2+2xy+3xz-4yz+2y^2-z^2$$
$$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))-4yz+2y^2-z^2$$
$$2(x^2+x(y+\frac{3}{2}z))+(\frac{1}{2}(y+\frac{3}{2}z))^2-(\frac{1}{2}(y+\frac{3}{2}z))^2-4yz+2y^2-z^2$$
Then things get kinda messy from here and I get totally lost from then on, could anyone help me out factoring this? And telling me if there is an eaay and non-confusing way to solve it? 

We can do it simply using $(a \pm b )^2 = a^2 \pm 2ab + b^2$
$$2x^2 + 2y^2 -z^2 + 2xy +3xz - 4yz $$
$$ x^2 + y^2 + 2xy + x^2 +2(x)(\frac{3}{2}z)+(\frac{3}{2}z)^2 -(\frac{3}{2}z)^2+y^2 -2(y)(2z)+(2z)^2 -(2z)^2-z^2$$
$$(x+y)^2 +(x+\frac{3}{2}z)^2 +(y-2z)^2 - (\frac{\sqrt{29}}{2}z)^2 $$
