(x+2\x-1)^2+6=5(x+2)\x-1?

Answer 1

I assume you want x.

To make this look like something you have seen, define the variable y=( x+2)/(x-1). Probably want to restrict x to something > 1. Doing this,
y^2 + 6 = 5y. We can the y-roots by solving:
y^2 -5y +6 = 0. The roots are y=2 and y=3.
Now go back to the definition to find x. There will be two allowable values.