Parallel Capacitors with different voltages
At the current state of our universe, charge is conserved. (This wasn't necessarily always the case. See this article on dark matter, for example, discussing the possibility that charged particles created shortly after the Big Bang lost their electric charge during the inflationary period.) This means that the sum of two relative charges held by the two capacitors before being connected to each other must be the same as the relative charge of the combined capacitor after being connected. When you place two capacitors in parallel, the total charge of the final system is the sum of the two original charges on the two earlier systems. In short, \$q_{_{tot}}=q_{_1}+q_{_2}\$.

Since \$q_{_{1}}=C_{_{1}}\cdot V_{_{1}}\$ and \$q_{_{2}}=C_{_{2}}\cdot V_{_{2}}\$ and also that
 \$q_{_{final}}=C_{_{final}}\cdot V_{_{final}}\$; and since we also know that when two capacitors are placed in parallel that the total system capacitance is the sum of the two original system capacitances, or \$C_{_{final}}=C_{_{1}}+ C_{_{2}}\$; it then follows directly that:

$$q_{_{final}}=q_{_1}+q_{_2}=C_{_{1}}\cdot V_{_{1}}+C_{_{2}}\cdot V_{_{2}}=\left(C_{_{1}}+ C_{_{2}}\right)\cdot V_{_{final}}$$

It's easy then to re-arrange this so that:

$$V_{_{final}}=\frac{C_{_{1}}\cdot V_{_{1}}+C_{_{2}}\cdot V_{_{2}}}{C_{_{1}}+ C_{_{2}}}$$

Note that the energy of the final system is not the sum of the energy of the original two systems:

$$\frac12 C_{_{final}}\,V_{_{final}}^{^2}\ne \frac12 C_{_{1}}\,V_{_{1}}^{^2}+\frac12 C_{_{2}}\,V_{_{2}}^{^2}$$

In fact, the total energy will be smaller by this factor:

$$\Delta W = W_{_{final}}-\left(W_{_1}+W_{_2}\right)=-\frac12 \left(V_{_1}-V_{_2}\right)^2\frac{C_{_{1}}\cdot C_{_{2}}}{C_{_{1}}+C_{_{2}}}$$

This is true regardless of how that difference is expended. You may want to read this hyperphysics page on where the energy loss goes when charging up a capacitor. This lost energy can be through realistic resistance dissipation as heat. But as you assume a smaller and smaller connection resistance, so that the current is very much larger and the transfer time much shorter, then an increasing amount of the lost energy goes into electromagnetic radiation.