Solve the following quadratic please. (HELP....)

x^2+2y^2=36
x+y= 6

... i don't get whole numbers for x or y so i think i am doing it wrong.
I should get whole numbers or it will simply be a little bit too complicated... (than expected anyway)

x^2 + 2y^2 = 36
x + y = 6
y = 6 - x
y^2 = 36 - 12x + x^2
Sub that back into the first equation:
x^2 + 2(36 - 12x + x^2) = 36
x^2 + 72 - 24x + 2x^2 = 36
3x^2 - 24x + 36 = 0
x^2 - 8x + 12 = 0
(x-6)(x-2) = 0
x = 6 or x = 2
y = 6 - x from before, so either x = 6 and y = 0 OR x = 2 and y = 4.

From equation 2 we have y = 6 - x

So substitute this into equation 2 to give
x^2 + 2(6-x)^2
= x^2 + 2( 36-6x-6x+x^2)
= x^2 + 72 -24x + 2x^2
= 3x^2 -24x + 72 = 36

So 3x^2-24x+36 = 0
We can simplify this by dividing by 3 to give
x^2 - 8x + 12 = 0

Factorize to give
(x - 6)(x - 2) = 0

Therefore x = 6 and y = 0
Or x = 2 and y = 4

y = 6 - x
x^2 + 2(6 - x)^2 = 36
x^2 + 2(36 - 12x + x^2) = 36
x^2 + 72 - 24x + 2x^2 = 36
3x^2 - 24x + 36 = 0
(3x - 18) (x - 2) = 0
3x = 18 OR x = 2
x = 6 OR x = 2

Substitute x = 6 into x + y = 6:
x = 6, y = 0

Substitute x = 2 into x + y = 6:
x = 2, y = 4

Then, y = 6 - x

x^2 + 2(6 - x)^2 = 36

that becomes x^2 - 8x + 12 = 0

Solutions are x1 = 6 & x2 = 2

Then, y1 = 0 & y2 = 4