Since the equation of the parabola is in its standard form then you can first find the x coordinate of the vertex like this:
[tex]\begin{gathered} \frac{-b}{2a}\Rightarrow\text{ x-coordinate} \\ \text{ Where} \\ y=ax^2+bx+c\Rightarrow\text{ standard form} \end{gathered}[/tex]
In this case, you have
[tex]\begin{gathered} y=-3x^2+12x-10 \\ a=-3 \\ b=12 \\ c=-10 \end{gathered}[/tex][tex]\frac{-b}{2a}=\frac{-12}{2(-3)}=\frac{-12}{-6}=2\Rightarrow\text{ x-coordinate}[/tex]
Now, replace this value into the equation of the parabola to find the y-coordinate of the vertex:
[tex]\begin{gathered} y=-3x^2+12x-10 \\ x=2 \\ y=-3(2)^2+12(2)-10 \\ y=-3(4)+24-10 \\ y=-12+14 \\ y=2 \end{gathered}[/tex]
Therefore, the vertex of the parabola is (2,2).
