Starting with what you very correctly did, since it is obvious that $a \neq 0$, we can reduce to common denominator and face the problem of finding the zero('s ?) of function
$$f(a)=2 (a-1) e^a-3 a^2+2$$ This is a transcendental equation (mixing exponential and polynomial terms) and, most of the time, there is no explicit solution except, sometimes, in terms of special function.
The first derivatives are
$$f'(a)=2 a \left(e^a-3\right)\qquad \text{and} \qquad f''(a)=2 (a+1) e^a-6$$ The first derivative cancels for $a=0$ and $a=\log(3)$. By the second derivative test $(f''(0)=-4)$ the first point corresponds to a maximum $(f(0)=0)$ and the second one $(f''(\log(3))=6 \log (3))$ to a minimum $(f(\log(3))=-4-3 \log ^2(3)+6 \log (3)<0)$. So, the root we look for is greater than $\log(3)$.
Taking all this information into account, we can now generate an estimate of the solution developing $f(a)$ as Taylor series around $a=\log(3)$. This gives
$$f(a)=-\left(4+3 \log ^2(3)-6 \log (3)\right)+3 \log (3) (a-\log (3))^2+O\left((a-\log
(3))^3\right)$$ which gives
$$a_0=\log (3)+\sqrt{\frac{4+3 \log ^2(3)-6 \log (3)}{3 \log (3)}}\approx 1.657$$ which is not to bad since, up to now, almost no calculus was used.
Now, we need to polish the root and Newton method will be simple. Starting with $a_0$ as defined above, the iterates will be
$$\left(
\begin{array}{cc}
n & a_n \\
0 & 1.657419 \\
1 & 1.569271 \\
2 & 1.555003 \\
3 & 1.554653
\end{array}
\right)$$
Edit just for your curiosity
In fact, the equation to be solved could rewrite
$$e^{-a}=\frac 23 \frac{(a-1)}{\left(a-\sqrt{\frac{2}{3}}\right) \left(a+\sqrt{\frac{2}{3}}\right)}$$ which has a formal solution in terms of the generalized Lambert function (have a look at equation $(4)$).
This is nice from a formal point of view but not practical at all.
