Answer
$(1,3),(1,-3)$
Work Step by Step
1. Use the 2nd equation and plugin the 1st to get $3x^2+9x=12$ or $x^2+3x-4=0$ or $(x+4)(x-1)=0$, thus $$x=-4, 1$
2. For $x=-4$, we have $y^2=-36$ no real solution
3. For $x=1$, we have $y^2=9$, thus $y=\pm3$
4. The solutions are $(1,3),(1,-3)$
