How many complex roots does a cubic equation have?
Since every real number can be thought of as a complex number, one way to answer this is to say the answer is 3, after "counting multiplicities". On the other hand, if you are after non-real complex roots, and if the coefficients of your cubic equation are real numbers, the answer will be 0 or 2.
A real number
If you are after non-real complex roots in a situation with real number coefficients, the same example above would provide a situation where there are 0 complex roots. Examples where there are two non-real complex roots are situations where they occur in complex conjugate pairs . For example,
It can be proved that if a polynomial has real coefficients, then the non-real complex roots of such a polynomial, if there are any, must occur in these complex conjugate pairs. That's why there can't be such cubics with 1 or 3 non-real complex roots.
One other thing worth mentioning: by the Intermediate Value Theorem, all cubics with real coefficients must have at least 1 real root (giving another reason such a cubic can't have 3 non-real complex roots).