To solve the exercise, replace the coordinates of the points into the given equation.
If after operating, the result is a true proposition then the point belongs to the graph of the equation.
If after operating, the result is a false statement then the point does not belong to the graph of the equation.
Point (A) (-3,6)
[tex]\begin{gathered} 3x^2+6y+5=-7 \\ 3(-3)^2+6(6)+5=-7 \\ 3\cdot9+36+5=-7 \\ 27+36+5=-7 \\ 65=-7 \\ \text{ False proposition} \end{gathered}[/tex]Therefore, the point is not on the graph of the equation.
Point (B) (-2,0)
[tex]\begin{gathered} 3x^2+6y+5=-7 \\ 3(-2)^2+6(0)+5=-7 \\ 3\cdot4+0+5=-7 \\ 12+5=-7 \\ 17=-7 \\ \text{ False proposition} \end{gathered}[/tex]Therefore, the point is not on the graph of the equation.
Point (C) (0,-2)
[tex]\begin{gathered} 3x^2+6y+5=-7 \\ 3(0)^2+6(-2)+5=-7 \\ 3\cdot0-12+5=-7 \\ 0-12+5=-7 \\ -7=-7 \\ \text{ True proposition} \end{gathered}[/tex]Therefore, the point is on the graph of the equation.
Point (D) (8,2)
[tex]\begin{gathered} 3x^2+6y+5=-7 \\ 3(8)^2+6(2)+5=-7 \\ 3\cdot64+12+5=-7 \\ 192+12+5=-7 \\ 209=-7 \\ \text{ False proposition} \end{gathered}[/tex]Therefore, the point is not on the graph of the equation.
Finally, the only point that is on the graph of the equation is point C.
Graphically
