To solve the equation \(2x-\sqrt{3x-1}=6\), we can start by isolating the square root term.

First, we can add \(\sqrt{3x-1}\) to both sides of the equation to get rid of the square root on the left side:

\[2x-\sqrt{3x-1}+\sqrt{3x-1}=6+\sqrt{3x-1}\]

Simplifying the left side, we have:

\[2x=6+\sqrt{3x-1}\]

Next, we can subtract 6 from both sides of the equation:

\[2x-6=6+\sqrt{3x-1}-6\]

Simplifying the right side, we have:

\[2x-6=\sqrt{3x-1}\]

Now, we can square both sides of the equation to eliminate the square root:

\[(2x-6)^2=(\sqrt{3x-1})^2\]

Simplifying both sides, we have:

\[4x^2-24x+36=3x-1\]

Rearranging the equation, we get:

\[4x^2-24x-3x+36+1=0\]

Combining like terms, we have:

\[4x^2-27x+37=0\]

This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. However, in this case, the quadratic equation does not factor nicely, and completing the square would be quite involved. Therefore, we will use the quadratic formula:

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

For our equation, \(a=4\), \(b=-27\), and \(c=37\). Substituting these values into the quadratic formula, we have:

\[x=\frac{-(-27)\pm\sqrt{(-27)^2-4(4)(37)}}{2(4)}\]

Simplifying, we get:

\[x=\frac{27\pm\sqrt{729-592}}{8}\]

\[x=\frac{27\pm\sqrt{137}}{8}\]

Therefore, the solutions to the equation \(2x-\sqrt{3x-1}=6\) are:

\[x=\frac{27+\sqrt{137}}{8}\] and \[x=\frac{27-\sqrt{137}}{8}\]