Your recurrence is correct. It’s first-order, so you really need only one initial value, and you might as well use $a(0)=0$. One way to solve it is with generating functions. Multiply the recurrence by $x^n$ and sum over $n\ge 0$:
$$\begin{align*}
\sum_{n\ge 0}a_nx^n&=\sum_{n\ge 0}\left(2a_{n-1}+n^2\right)x^n\\
&=\sum_{n\ge 0}2a_{n-1}x^n+\sum_{n\ge 0}n^2x^n\\
&=2x\sum_{n\ge 0}a_{n-1}x^{n-1}+\sum_{n\ge 0}n^2x^n\\
&=2x\sum_{n\ge 0}a_nx^n+\sum_{n\ge 0}n^2x^n\;,
\end{align*}$$
where $a_{-1}=0$; note that the recurrence is then valid at $n=0$. If $g(x)=\sum_{n\ge 0}a_nx^n$, we now have
$$g(x)=2xg(x)+\sum_{n\ge 0}n^2x^n\;.$$
To find a closed form for $\sum_{n\ge 0}n^2x^n$, start with $\sum_{n\ge 0}x^n=\frac1{1-x}$. Differentiate both sides with respect to $x$, multiply both sides by $x$, differentiate again, and multiply again by $x$, and you’ll find (if I’ve made no mistakes) that
$$\sum_{n\ge 0}n^2x^n=\frac{x(1+x)}{(1-x)^3}\;.$$
Thus,
$$g(x)=\frac{x(1+x)}{(1-x)^3(1-2x)}\;.$$
If you now decompose the righthand side into partial fractions, you’ll find that each has a straightforward series expansion; for example,
$$\frac{a}{(1-x)^2}=a\sum_{n\ge 0}(n+1)x^n\;.$$
You can combine these expansions into a single power series and equate coefficients to get a closed form for $a_n$. I’ll leave the details to you, but I get $$a_n=6\cdot2^n-n^2-4n-6\;.$$
