Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 12

$(0,-2),(8,6)$

Substituting the second equation ($y=x-2$) into the first one we get: $-x^2+2(x-2)^2=8\\-x^2+2(x^2-4x+4)=8\\x^2-8x+8=8\\x(x-8)=0$ Thus $x=0$ or $x=8$. If $x=0$, then $y=-2$, and if $x=8$, then $y=6$. Thus the solutions are: $(0,-2),(8,6)$