Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Quiz for Lessons 9.6-9.7 - Page 664: 9

$(2,0),(-2,-8)$

Plugging the transformed version of the second equation ($y=2x-4$) into the first one we get: $x^2+4x+(2x-4)^2+6(2x-4)=12\\5x^2-8=12\\5x^2-20=0\\x^2=4$ Thus $x=2$ or $x=-2$. $y=2x-4$, thus if $x=2$, then $y=0$, and if $x=-2$, then $y=-8$. Thus the solutions are: $(2,0),(-2,-8)$