First, let's solve the second equation for x. We have x + y = -4, so if we isolate x, we get x = -4 - y.
Now, substitute this value of x into the first equation: (-4 - y)^2 + y^2 = 40.
Expanding and simplifying the equation, we get: 16 + 8y + y^2 + y^2 = 40.
Combining like terms, we have: 2y^2 + 8y - 24 = 0.
Now, let's solve this quadratic equation. We can factor it as: 2(y + 4)(y - 3) = 0.
Setting each factor equal to zero, we have two possible solutions: y + 4 = 0 or y - 3 = 0.
Solving for y in each case, we get y = -4 or y = 3.
Now, substitute these values of y back into the second equation x + y = -4 to find the corresponding values of x.
When y = -4, we have x + (-4) = -4, which gives us x = 0.
When y = 3, we have x + 3 = -4, which gives us x = -7.
So, the solutions to the system of equations are (x, y) = (0, -4) and (-7, 3).