Given
[tex]1.33x^2-16x-16y=-64[/tex]Transform it into its vertex form as shown below
[tex]\begin{gathered} \Rightarrow16y=1.33x^2-16x+64 \\ \Rightarrow y=\frac{1.33}{16}x^2-x+4 \end{gathered}[/tex]Complete the square on the right side of the equation,
[tex]\begin{gathered} \\ \\ Set\text{ }b\text{ the term that completes the square} \\ \\ \\ \Rightarrow ax^2-2abx+ab^2=\frac{1.33x^2}{16}-x+c \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow a=\frac{1.33}{16} \\ -2abx=-x \\ \Rightarrow2ab=1 \\ \Rightarrow b=\frac{1}{2a} \\ \Rightarrow b=\frac{16}{2*1.33} \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} \Rightarrow y=\frac{1.33}{16}x^2-x+4=\frac{1.33}{16}(x-\frac{8}{1.33})^2-(\frac{8}{1.33})^2+4 \\ \Rightarrow y=\frac{1.33}{16}(x-\frac{8}{1.33})^2-((\frac{8}{1.33})^2-4) \end{gathered}[/tex]However, consider that the leading coefficient is 1.33=1/3
Then, the given expression becomes
[tex]\begin{gathered} \frac{1}{3}x^2-16x-16y=-64 \\ \end{gathered}[/tex]By the same token,
[tex]\begin{gathered} \Rightarrow y=\frac{1}{48}x^2-x+4 \\ \end{gathered}[/tex]Completing the square,
[tex]\frac{1}{48}(x^2-48x)+4=\frac{1}{48}(x-24)^2-8[/tex][tex][/tex]