To solve this system of equations, we can use substitution method or elimination method.

Let's solve it using substitution method:

1) From the second equation, we can express x as x = 16 - 4y
2) Substitute x in the first equation:
2(16-4y) + 2y^2 = 32
32 - 8y + 2y^2 = 32
2y^2 - 8y = 0
2y(y-4) = 0
y = 0 or y = 4

If y = 0:
x = 16 - 4(0) = 16
Therefore, the solution is (16, 0)

If y = 4:
x = 16 - 4(4) = 16 - 16 = 0
Therefore, the solution is (0, 4)

So, the system of equations has two solutions: (16, 0) and (0, 4).