Finding all truth assignment to 2SAT I'm looking for a way to assign true/false value to variable A,B,C,D so that all clauses are satisfies. 
Formu
(A ∨ ¬B) ∧ (¬A ∨ ¬C) ∧ (A ∨ B) ∧ (¬C ∨ D) ∧ (¬A ∨ D).

For that, A=True, B=False, C=False, D=True satisfy the all clauses. What's the best approach? I'm using truth table to figure this out but I'm wondering if there is better way.
Edit : Seems that the question is very confusing for many people. So I have edited it.
 A: Construct a graph where each variable and its negation are nodes in this graph. If there are $n$ variables, this should be a graph with $2n$ nodes.
Note that $A\lor B$ is equivalent to $\lnot B\implies A$ and $\lnot A\implies B$.
Hence each clause in the 2-SAT would give 2 edges.
If the graph has a variable and negation in the same strongly connected component, then the 2-SAT is unsatisfiable.
Using the graph, if $A\implies\lnot A$, it means that $\lnot A$.
If a variable cannot be deduced from such, you can arbitrarily set it true or false to generate all possibilities.
A: I'm not sure what the question is exactly, but if it is the one as in your title, counting solutions to 2-SAT is computationally hard. More specifically, this problem is known as #2-SAT, and it is #P-complete, see e.g. here. 
However, it is possible to count the number of solutions in $O(1.264^n)$ time, where $n$ is the number of variables.
A: This particular instance in the OP can be solved just by inspection. There are only $4$ variables, so there are only $2^4=16$ possible assignments. And it is straightforward to quickly narrow-down further.

*

*Here you can see that, in the instance in the OP, $A$ must be positive [the only way both $(A \vee B)$ and $(A \vee \neg B)$ can both be satisfied].


*Then $C$ must be negative [the only way $(\neg A \vee \neg C)$ can be satisfied, recall we just found that $A$ must be positive]. Furthermore, by similar reasoning, so must $D$ be positive.


*Then $B$ can be either positive or negative.


*This comes to exactly two solutions $(A, \neg B, \neg C, D)$, and $(A,B,\neg C,D)$.
For general instances, the algorithm specified by Element118 will do to find a solution should one exist, but it is not really needed here.
