In the {\sc Min-Sum 2-Clustering} problem, we are given a graph and a parameter $k$, and the goal is to determine if there exists a 2-partition of the vertex set such that the total conflict number is at most $k$, where the conflict number of a vertex is the number of its non-neighbors in the same cluster and neighbors in the different cluster. The problem is equivalent to {\sc 2-Cluster Editing} and {\sc 2-Correlation Clustering} with an additional multiplicative factor two in the cost function. In this paper we show an algorithm for {\sc Min-Sum 2-Clustering} with time complexity $O(n\cdot 2.619^{r/(1-4r/n)}+n^3)$, where $n$ is the number of vertices and $r=k/n$. Particularly, the time complexity is $O^*(2.619^{k/n})$ for $k\in o(n^2)$ and polynomial for $k\in O(n\log n)$, which implies that the problem can be solved in subexponential time for $k\in o(n^2)$. We also design a parameterized algorithm for a variant in which the cost is the sum of the squared conflict-numbers. For $k\in o(n^3)$, the algorithm runs in subexponential $O(n^3\cdot 5.171^θ)$ time, where $θ=\sqrt{k/n}$.