What will happen if the potential is less than 0, for instance $V(x)=-10eV$. Is this means there will be no bound states? Since solution to the time independent Schrodinger equation (those discrete energies) must be greater than 0.
Your finite square well potential looks like:
where $V_0$ is the potential energy outside the well and $V_1$ is the potential inside the well. The depth of the well is $\Delta V = V_0 - V_1$.
We normally take $V_0$ to be zero, in which case $V_1$ is negative (like your $-10$eV) and $\Delta V = V_1$. However you can add any constant value to the potential energies without changing the physics. That's because the wavefunction only depends on the well depth $\Delta V$, and adding the same constant term to $V_0$ and $V_1$ doesn't change $\Delta V$. We say the potential energy has a global gauge symmetry.
It's not clear from your question what you intended to have the value $-10$eV. You could set $V_0 = -10$eV and particles would still be bound in the well as long as $V_1$ was less than $-10$eV, that is $V_1 \lt V_0$.