Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 15

$(-1,-4),(-6.5,7)$

Substituting the second equation ($y=-6-2x$) into the first one we get: $4x^2-5(-6-2x)^2=-76\\4x^2-5(36+24x+4x^2)=-76\\-16x^2-104-120x=0\\2x^2+13+15x=0\\(x+1)(2x+13)=0$ Thus $x=-1$ or $x=-6.5$. If $x=-1$, then $y=-4$, and if $x=-6.5$, then $y=7$. Thus the solutions are: $(-1,-4),(-6.5,7)$