We have the following:
[tex]\begin{gathered} y=x^2-3x+6 \\ y=2x+6 \end{gathered}[/tex]We subtract the equations:
[tex]\begin{gathered} y-y=x^2-3x+6-2x-6 \\ 0=x^2-5x \\ 0=x(x-5) \\ x=0;x=5 \end{gathered}[/tex]for y:
[tex]\begin{gathered} y=2\cdot0+6 \\ y=6 \\ y=2\cdot5+6 \\ y=16 \end{gathered}[/tex]therefore, the answer is:
(0,6) and (5,16), the option D.
