Conjugate, non-conjugate and single poles in an RLC circuit system
The circuit you have drawn, the \$RLC\$ network, does not feature zeros, only poles. Should you add a small resistance in series with the capacitor, then you would add a zero. The equation you have derived shows a second-order polynomial form in the denominator: \$D(s)=1+b_1s+b_2s^2\$. It can be advantageously factored in a canonical form such as \$D(s)=1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2\$ in which \$Q=\frac{\sqrt{b_2}}{b_1}\$ and \$\omega_0=\frac{1}{\sqrt{b_2}}\$. By quick identification with your expression, you can determine these values easily. Now, if you try to determine the roots of \$D(s)=0\$, or the poles of the transfer function, you end up with a pair of roots defined by \$s_{p1},s_{p2}=\frac{\omega_0}{2Q}(\pm\sqrt{1-4Q^2}-1)\$. 

From this expression, you can see that depending on the value of \$Q\$, the expression under the square root can be positive or negative. If positive, the roots are real (there is no imaginary part) and the transient response is non-oscillatory. This is the case for a \$Q\le0.5\$. For a very low \$Q\lt\lt1\$, we can apply the so-called low-\$Q\$ approximation in which you consider the poles well spread apart. There is one in the low frequency domain (the root in your diagram is close to the vertical axis) while the second is in higher frequencies. You thus write \$D(s)\approx(1+\frac{s}{\omega_p1})(1+\frac{s}{\omega_p2})\$ in which \$\omega_{p1}=Q\omega_0\$ and \$\omega_{p2}=\frac{\omega_0}{Q}\$. It simply means that for a high value of the series resistance, the circuit can be replaced by two (isolated) cascaded \$RC\$ networked tuned at \$\omega_{p1}\$ and \$\omega_{p2}\$.  If you drive your circuit with a step voltage, the output voltage is very sluggish without overshoot.

When \$Q=0.5\$, the roots are still real but coincident. The response is still non-oscillatory and \$D(s)\approx(1+\frac{s}{\omega_0})^2\$. If you drive your circuit with a step voltage, the output voltage rises quickly without overshoot.

Now if you reduce the series resistance further, \$Q\$ exceeds 0.5 and the roots involve imaginary notation: they become imaginary conjugates like one root located at -1+2j and the other at -1-2j for instance. Please note that the real part in these expressions is negative implying that in your \$s\$-plane diagram, the little "crosses" are in the left half-plane (LHP). We call them LHP poles or LHPP. The negative real parts model the losses in your circuit, the power dissipation that will damp oscillations and force the time-domain response to converge to a steady-state value as \$t\$ approaches infinity. If you drive your circuit with a step voltage, the output voltage rises quickly and overshoots.

As these losses become less and less (the circuit gains in efficiency), \$Q\$ increases - more pronounced overshoot in \$v_{out}(t)\$ - until the poles become pure imaginaries: the real parts are gone and the poles are located on the vertical axis. Should you excite the circuit with a transient pulse, the response would be oscillatory with sustained oscillations when the excitation is gone.

A pole at the origin is a different story. Having a pole at the origin simply means that your transfer function hosts a division by \$s\$: \$H(s)=\frac{1}{s(1+\frac{s}{\omega_p})}\$ features a LHP pole and a pole at the origin because \$D(s)=0\$ for \$s=0\$. Should you divide by \$s^2\$ instead, you have two poles at the origin etc. You have a pole at the origin in systems featuring an integrator for instance. The pole at the origin implies a gain approaching infinity in dc (when \$s=0\$), actually the op amp open-loop gain, which ensures a very low static error in the variable you control.