Chapter 9 - Quadratic Functions and Equations - 9-8 Systems of Linear and Quadratic Equations - Practice and Problem-Solving Exercises - Page 600: 36

The solutions of the system are $(3,18)$ or $(-1,6)$.

The first equation: $y=x^2+x+6$ The second equation: $y=2x^2-x+3$ Set two equations equal: $x^2+x+6=2x^2-x+3$ $x^2-2x-3=0$ $(x-3)(x+1)=0$ $x-3=0$ or $x+1=0$ $x=3$ or $x=-1$ Solve for y: $y=(3)^2+3+6$ or $y=(-1)^2+(-1)+6$ $y=18$ or $y=6$ The solutions of the system are $(3,18)$ or $(-1,6)$.