# $f(x)=x^2-3x-2$ for $x\ge3$. how to find point of intersection of $f^{-1}(x)$ and $g(x)=x-3$?

We have $$f(x)=x^2-3x-2$$ where $$x\ge3$$. graph of two functions $$y=f^{-1}(x)$$ and $$g(x)=x-3$$ intersect each other at the point $$A$$. What is the distance from the point $$A$$ to the origin?

$$1)\sqrt{74}\qquad\qquad\qquad2)\sqrt{69}\qquad\qquad\qquad3)\sqrt{89}\qquad\qquad\qquad4)\sqrt{97}$$

It is a problem from a timed exam, so I am looking for the fastest answers.

One approach is completing the square of $$x^2-3x-2$$ then find $$f^{-1}(x)$$ and equate it with $$g(x)=x-3$$, but since completing the square containing some fractions it seems not a good idea to solve the problem quickly.

Another approach: I noticed that $$f(x)$$ and $$y=x$$ intersect each other at the same point(s) as $$f^{-1}(x)$$ intersects $$y=x$$ (because $$f$$ and $$f^{-1}$$ are symmetrical along $$y=x$$) so we can find the points of intersections of $$f^{-1}(x)$$ and $$y=x$$ by solving the equation $$x^2-3x-2=x$$, but I can't make a connection between those points and points of intersections of $$f^{-1}(x)$$ and $$y=x\color{red}{-3}$$.

The inverse function is given by $$y$$ where $$x=y^2-3y-2$$ (just swap the x and the y).
Line is $$y=x-3$$
Write that as $$x=y+3$$ and equate the two;
$$y^2-3y-2=y+3$$