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 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 and from wikipedia.
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?
