3d particle in a constant potential radial equation

For the free particle in 3d, I follow Robinett‘s page 491 to find that the solutions to the radial equation are the spherical Bessel function $j_l(z)$ and $y_l(z)$ where $z=kr$ and $k=sqrt{frac{2mE}{hbar^2}}$.

If I allow for a constant potential $V=V_0$, and consider the $E<V$ case, then since the state are bound, I only use the regular Bessels $j_l(z)$.

I am struggling to figure out the other things that change in the solution. Wikipedia state the following:

$$ R(r) = A j_l (z) $$ with

$$ z= sqrt{frac{2m(E-V_0)}{hbar^2}} r $$ and $A$ is some normalization.

1. Why is it $E$ MINUS $V_0$? if $E<V$ then this is negative and the function is imaginary from the $sqrt{-1}$ no?
   Wiki is unclear on this, since they state

"We first consider bound states, i.e., states which display the particle mostly inside the box (confined states). Those have an energy E less than the potential outside the sphere, i.e., they have negative energy…"

but then directly below the $R(r)$ solution there is a contradiction with this:

"Note that for bound states, $ V_{0}<E<0$"

2. How is the normalization constant A solved for from the boundary conditions?