Suppose I have a "quadratic equation" whose coefficients are functions of the variable to be solved for:
$$f(x)x^2+g(x)x+h(x)=0,$$ with $f,g,h\neq 0$.
Would it make sense to apply the quadratic formula to solve for $x$ here? I realize the answer will be $$x=u(x),$$ for some generalised function $u$, but my question is would this equality be correct?
You can complete the square in one of the usual ways - multiply by $4f(x)$ to conclude $$4f(x)^2x^2+4f(x)g(x)x+4f(x)h(x)=0$$ which gives $$(2f(x)x+g(x))^2=g(x)^2-4f(x)h(x)$$
Then take the square root (which requires some care) and rearrange - the process works. What it achieves depends on what the various functions are.