Since the equation of the parabola is in its standard form then you can first find the x coordinate of the vertex like this:
![\begin{gathered} (-b)/(2a)\Rightarrow\text{ x-coordinate} \\ \text{ Where} \\ y=ax^2+bx+c\Rightarrow\text{ standard form} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ko840pj6joftvug0e4e8x3jz07gxur8jbq.png)
In this case, you have
![\begin{gathered} y=-3x^2+12x-10 \\ a=-3 \\ b=12 \\ c=-10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/22v5lbvl7db8xnoc326wfa2irbbjoh3wxo.png)
![(-b)/(2a)=(-12)/(2(-3))=(-12)/(-6)=2\Rightarrow\text{ x-coordinate}](https://img.qammunity.org/2023/formulas/mathematics/college/hro840kijdh2ye3j057pphtc76trtlhfb8.png)
Now, replace this value into the equation of the parabola to find the y-coordinate of the vertex:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/gs2ut83x6mxj1r3l89kskic20gfeuk0znm.png)
Therefore, the vertex of the parabola is (2,2).