Propositional formula in DNF can be decided in polynomial time?

For a given propositional formula f in DNF, one can decide in polynomial time, if the formula is satisfiable: Just walk through all subformulas (l_1 and ... and l_k) and check, wheter it has NO complementary pair of literals. Formula f is satisfiable iff such subformula exists.

Is my approach above correct ?

If yes, I'm wondering why all modern SAT solvers get a CNF as input format, and don't just use DNF.

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The conversion from CNF to DNF can come at an exponential cost. For example $(a_1 \lor b_1) \land \cdots \land (a_n \lor b_n)$ expands to $2^n$ many terms. As you comment, for DNF satisfiability is easy - it is falsifiability which is hard. If the problem is trivial, you don't input it to a SAT solver, and that's why SAT solvers accept CNFs instead of DNFs.