Question: The 3-SAT problem is NP-complete, meaning that no known algorithm can provide an exact solution in polynomial time, while a solution can be tested very quickly in polynomial time. 

My question is, if asking for an algorithm that only provides yes/no, or solvable/not solvable, without providing an exact solution of the 3-SAT formula, will this result in another complexity class? 

The reason I have in mind of asking this, is a similar situation when looking for prime numbers and factorizations. Where it is known that it's much easier to test if a number is a prime number yes/no, than to find an exact prime factorization of a number.

Answer: 
Check Belare and Goldwasser, "The complexity of decision versus search", SIAM J. Of Computing, 23:1 (feb 1994), pp. 97-119. Belare has [notes for a class](https://cseweb.ucsd.edu/~mihir/cse200/decision-search.pdf).
Short answer: if you can solve the decision (yes/no) problem, calling that tells you if it has no solution; if there is a solution, pick a variable and set it to true, see if the result can be satisfied; if not, it has to be false. This way, with one call to the oracle per variable you get a "solution" (set of values of the variables making the expression true, it here are any). Thus the cost of the search problem isn't "too much more" (multiply by the size of the input --number of variables-- is polynomial).