I think you got the fusion reaction slightly wrong, what is happening *in the end* is

\begin{eqnarray}
^{2}\mbox{H} +\ ^{2}\mbox{H} &\rightarrow& ^{3}\mbox{H} +\ ^{1}\mbox{H}\\
^{2}\mbox{H} +\ ^{2}\mbox{H} &\rightarrow& ^{3}\mbox{He} + \mbox{n},
\end{eqnarray}

which both have approximately the same probability to occur. Note that the released kinetic energy is *not* included in the reactions above.

Insert the [atomic masses](https://en.wikipedia.org/wiki/Atomic_mass) results in

\begin{eqnarray}
2.01410177812\cdot u + 2.01410177812\cdot u  &=& 3.0160492779\cdot u + 1.00782503223\cdot u + x_1\\
2.01410177812\cdot u + 2.01410177812\cdot u  &=& 3.0160293201 \cdot u + 1.00866491588\cdot u + x_2
\end{eqnarray}

where $x_{1,2}$ corresponds to the released energy. The data for the atomic masses is from the [NIST website](http://www.nist.gov/pml/data/comp.cfm) and from [wikipedia](https://en.wikipedia.org/wiki/Neutron).

Solving the equatione yields 

\begin{eqnarray}
x_1 &\approx& 0.00432924611\cdot u\\
x_2 &\approx& 0.00350932026\cdot u
\end{eqnarray}

corresponding to ($1\cdot u$ corresponds approximately to $931.5$ MeV)

\begin{eqnarray}
x_1 &\approx& 4.03\,\mathrm{MeV}\\
x_2 &\approx& 3.27\,\mathrm{MeV}.
\end{eqnarray}

So this is the *released energy* for both reactions, does the reaction ''equation'' now makes more sense to you?