Here is my attempt for a $d$-dimensional $V$ over an algebraically closed field, like $\mathbb{C}$. Let's identify $\text{Sym}^2(V)$ with the space of all symmetric, contravariant, rank-$2$ tensors or identically the space of all degree $2$ homogeneous polynomials in $d$ variables $(x_1, \cdots , x_d) \equiv \mathbf{x}$. Moreover, we know that every $1$-parameter subgroup $\lambda: \mathbb{C}^* \rightarrow G$ of $G=SL(V)$ is conjugate to
\begin{equation}
\mathbb{C}^* \ni t \mapsto \lambda(t) =
\begin{bmatrix}
t^{a_1} & & \\
& \ddots & \\
& & t^{a_d} \\
\end{bmatrix},
\end{equation}
for some integers $a_i \in \mathbb{Z}$, such that $a_1 \geq \cdots \geq a_d$ and $\sum_{i=1}^da_i = 0$. According to Hilbert-Mumford criterion, a polynomial $f (\mathbf{x})= \sum_{i,j}c_{ij}x_ix_j$ is unstable if and only if there exists a 1-parameter subgroup $\lambda$ of $SL(V)$ such that $\lim_{t\rightarrow 0} \lambda(t) \, . \, f (\mathbf{x}) = 0$, where
\begin{equation}
\lambda(t) \, . \, f (\mathbf{x})= f(\lambda^{-1}(t) \, . \, \mathbf{x}) =\sum_{i,j} c_{ij} t^{-a_i - a_j}x_ix_j,
\end{equation}
which then implies that $f$ in unstable if and only if there exists a $\lambda$ of $SL(V)$ such that $a_i + a_j <0$ for every term in the sum. This means that $f$ can be destabilized ($\lambda(t) \, . \, f$ tends to $0$ as $t \rightarrow 0$). For generalizations to other fields, you may want to take a look at the definition (4.3) or remark (4.5) here.