1. **State the problem:** Find the distance from the point $(2,5)$ to the circle given by the equation $$x^2 + y^2 - 6x - 4y - 3 = 0.$$\n\n2. **Rewrite the circle equation in standard form:**\nComplete the square for $x$ and $y$ terms.\n\n$$x^2 - 6x + y^2 - 4y = 3$$\n\nFor $x$: $$x^2 - 6x = (x^2 - 6x + 9) - 9 = (x - 3)^2 - 9$$\nFor $y$: $$y^2 - 4y = (y^2 - 4y + 4) - 4 = (y - 2)^2 - 4$$\n\nSubstitute back: $$ (x - 3)^2 - 9 + (y - 2)^2 - 4 = 3 $$\n\nSimplify: $$ (x - 3)^2 + (y - 2)^2 - 13 = 3 $$\n$$ (x - 3)^2 + (y - 2)^2 = 16 $$\n\nSo the circle has center $C = (3, 2)$ and radius $r = 4$.\n\n3. **Find the distance from point $P = (2,5)$ to the center $C$: $$d = \sqrt{(2 - 3)^2 + (5 - 2)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}.$$**\n\n4. **Distance from point to circle:**\nThe distance from point $P$ to the circle is the absolute difference between $d$ and $r$: $$\text{distance} = |d - r| = |\sqrt{10} - 4|.$$\n\n5. **Final answer:**\n$$\boxed{4 - \sqrt{10}}$$ (since $4 > \sqrt{10}$, the point lies outside the circle).