Why do we know mu?
It was non-obvious to me why we pick the quadratic root with '-' instead of the one with '+'. Here is some reasoning that helped: We want T to point down. Vertical component of "mu I" is pointing down (and has magnitude "mu cos(theta i)"). "gamma N" could be up or down.
Case 1: "gamma N" is down, then gamma is negative ==> gamma =/= mu cos(theta i) + cos(theta t).
Case 2: "gamma N" is up. Then we want vertical component of "mu I" larger than magnitude of "gamma N" so that T is down ==> mu cos(theta i) > gamma ==> gamma =/= mu cos(theta i) + cos(theta t).
@jeyla PBR book calculates it in more steps by breaking up the vectors into horizontal and vertical components. One way to think of it: T's horizontal component must come from "mu I" term since "gamma N" is vertical.
Magnitude of horizontal component of I is sin(theta i) and magnitude of horizontal component of T is sin(theta t). Their ratio is sin(theta t) / sin(theta i) = ni / nt (which matches the mu picked on this slide).