Your f(x)=(x+5)^4 +k expression
defines what f(x) is, but does not have a "solution" per se. No matter what value of k one chooses, if you pick an x, one can compute an f(x).
Something has been omitted from your question that needs to be there. Perhaps they are lookig for "solutions" to f(x) = 0. If so, the question should have said so.
In the function f(x)=(x+5)^4 +k , for which values of k does the function have two real solutions? no real solutions? Explain?
2 answers
Graph is same as y=x^2, minimum is (-5,k), if k>0 then no real solution; if k<0 then two real solution; if k=0 then four real solution (four same real solutions)