Subtracting x from the given equation we get:
[tex]6x^2-x=2+x-x=2.[/tex]
Dividing by 6 we get:
[tex]x^2-\frac{1}{6}x=\frac{1}{3}\text{.}[/tex]
Notice that:
[tex]x^2-\frac{1}{6}x=x^2+2(1)(-\frac{1}{12})x\text{.}[/tex]
Therefore:
[tex]x^2+2(1)(-\frac{1}{12})x=\frac{1}{3}[/tex]
Adding (-1/12)² from the above equation we get:
[tex]\begin{gathered} x^2+2(1)(-\frac{1}{12})x+(-\frac{1}{12})^2=\frac{1}{3}+(-\frac{1}{12})^2, \\ (x-\frac{1}{12})^2=\frac{1}{3}+\frac{1}{144}, \\ (x-\frac{1}{12})^2=\frac{49}{144}. \end{gathered}[/tex]
Solving for x we get:
[tex]\begin{gathered} (x-\frac{1}{12})^2=\frac{7^2}{12^2}, \\ x-\frac{1}{12}^{}=\pm\frac{7^{}}{12^{}}, \\ x=\frac{1}{12}\pm\frac{7^{}}{12^{}}, \\ x=\frac{2}{3}\text{ or x=-}\frac{1\text{ }}{2}\text{.} \end{gathered}[/tex]
Answer: Completing the square gives us:
[tex](x-\frac{1}{12})^2=\frac{49}{144}.[/tex]
