Hello, thank you for posting your question!
An extraneous solution would mean one where it solves the underlying quadratic equation but doesn’t actually fit into the original equation.
So I would start on this one by squaring both sides to set up the quadratic! Then once you get the values for that you can go back and plug them back in.
It should end up being x = 7/4 and x = -1 for the two values that solve the quadratic ... then if you go back to the original function and plug those values back in, x = -1 ends up being extraneous because it doesn't logically hold for the original equation.
I hope that helps get you moving in the right direction! Feel free to reach out if you have any more questions beyond that :)