Yup, a fourth-degree polynomial has exactly four roots (always considering multiplicity).
An nth-degree polynomial has exactly n roots.
The key to this exercise is recognizing that the given polynomial is quadratic "in form".
We handle this by making a substitution (at the beginning), and then reversing the substitution (at the end).
Let z = x^2
Now make the substitution.
z^2 - 50z + 49 = 0
This 2nd-degree polynomial equation has exactly two solutions.
After you find them, substitute back.
In other words, set each of the two solutions equal to x^2.
The resulting two equations each have two solutions, giving a total of four solutions to the original equation.
Does that help?