Answer
$\{(2+2\sqrt 2\ i, 5-\sqrt 2\ i), (2-2\sqrt 2\ i, 5+\sqrt 2\ i)\}$
Work Step by Step
1. From the second equation, we have $2y=12-x$
2. Use the above relation in the first equation to get $12-x=3x-x^2$ or $x^2-4x+12=0$
3. Use the quadratic formula, we have $x=\frac{4\pm\sqrt {4^2-4(12)}}{2}=2\pm2\sqrt 2\ i$
4. We can find the corresponding $y$ as $y=5\mp\sqrt 2\ i$
5. Thus the solution set is $\{(2+2\sqrt 2\ i, 5-\sqrt 2\ i), (2-2\sqrt 2\ i, 5+\sqrt 2\ i)\}$
