For simple rockets, using the rocket equation is simple enough.
$$ \Delta v = v_e ln\left(\frac{m_i}{m_f}\right) = I_{sp} \ g \ ln\left(\frac{m_i}{m_f}\right) $$
plug in full and empty weights and Isp, and you're done.
But what to do when a rocket uses boosters? You get 2 Isp values. I expect you can't just add the delta-V of the main stages to the delta-V of the individual boosters.
Is there a way to use the Rocket equation on rockets that use boosters?
For each phase of flight (stage or throttle variation or what have you) you weight the specific impulse (= exhaust velocity) by the propellant mass flow rates (i.e. consumption rates) of the different engines operating.
So if, for example the boosters are consuming 5000kg/s at 240s specific impulse and the core is consuming 1000kg/s at 310s, it works out to
$$ \frac {5000 \frac {kg} {s} * 240 {s} + 1000 \frac {kg} {s} * 310{s} } {5000 \frac {kg} {s} + 1000 \frac {kg} {s} } = 251.66 {s} $$
For 251.66s specific impulse.
You have to consider each "phase" of flight.
For each phase of flight you need to work out the initial mass, the final mass and the overall specific impulse (total thrust/total propellant flow rate). You can then apply the rocket equation to each phase of the flight and add up the results.
The derivation is based on the rate of change of total momentum
$$\frac{dp_{tot}}{dt}$$.
For each engine,
$$\frac{dp}{dt}=v_{ex}\frac{dm}{dt},$$
the exhaust velocity times the rate that the propellants' mass is ejected (kg/sec). You calculate that for each engine, then add them up.
So where you would normally have
$$m \frac{dV}{dt} = -v_{ex}\frac{dm}{dt},$$
you would have instead
$$m \frac{dV}{dt} = -\Sigma \left(v_{ex, i}\frac{dm_i}{dt}\right),$$
where $v_{ex,i}$ and $\frac{dm_i}{dt}$ are the exhaust velocity and mass rate of the ith engine. You can then introduce the weighted exhaust velocity:
$$ v_{ex,m} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\sum \frac{dm_i}{dt}} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\frac{dm}{dt}}$$
Hence you have
$$m\frac{dV}{dt} = -v_{ex,m}\frac{dm}{dt}$$
which is the same equation as in the single engine case. Integrating this equation yields the following rocket equation:
$$\Delta V = v_{ex,m} ln\big(\frac{m_i}{m_f}\big)$$
uhoh's answer shows how to derive combined ISP from momentum and rate. Here is the practical application of that, using the list of variables that's usually available:
When the engines are burning together, we can get the combined average exhaust velocity the same way from trust and flow rate:
$$v_{avg} = \frac{F}{\frac{dm}{dt}} = \frac{F_c + F_b}{\frac{dm_c}{dt}+\ \frac{dm_b}{dt}} = \frac{F_c + F_b}{\frac{F_c}{v_c} + \frac{F_b}{v_b}}$$
The burn will be in two parts. First the core and boosters burning together, and then the boosters will be detached and the core use its remaining propellant.
First, we find the mass flow of the boosters:
$$\frac{dm_b}{dt_b} = \frac{F_b}{v_b}$$
From that, we can find the booster burn time:
$$T_{boosters} = \frac{v_b \left(M_{b0} - M_{b1}\right)}{F_c}$$
From which we can find how much mass the core has used in the same time:
$$M_{core\ partial} = \frac{F_c}{v_c} \cdot T_{boosters}$$
$$M_{core\ partial} = \frac{v_b \left(M_{b0} - M_{b1}\right)}{v_c}$$
We now have the mass ratio and the combined average exhaust velocity for the first burn:
$$\Delta v_1 = v_{avg} \cdot \ln{\left(\frac{M_p + M_{c0} + M_{b0}}{M_p + M_{c0} - M_{core\ partial} + M_{b1}}\right)}$$
And for the second part, just the core alone, with the remaining propellant:
$$\Delta v_2 = v_{c} \cdot \ln{\left(\frac{M_p + M_{c0}- M_{core\ partial}}{M_p + M_{c1}}\right)}$$
Or combined and expanded:
$$\Delta v_{total} = \Delta v_1 + \Delta v_2$$
$$\Delta v_{total} = v_{avg} \cdot \ln{\left(\frac{M_p + M_{c0} + M_{b0}}{M_p + M_{c0} - M_{core\ partial} + M_{b1}}\right)} + v_{c} \cdot \ln{\left(\frac{M_p + M_{c0}- M_{core\ partial}}{M_p + M_{c1}}\right)}$$
$$\Delta v_{total} = \frac{F_c + F_b}{\frac{F_c}{v_c} + \frac{F_b}{v_b}} \cdot \ln{\left(\frac{M_p + M_{c0} + M_{b0}}{M_p + M_{c0} - \frac{v_b \left(M_{b0} - M_{b1}\right)}{v_c} + M_{b1}}\right)} + v_{c} \cdot \ln{\left(\frac{M_p + M_{c0}- \frac{v_b \left(M_{b0} - M_{b1}\right)}{v_c}}{M_p + M_{c1}}\right)}$$
