The value(s) of $y$ for which the following pair of equations $x^2+y^2+16=0\text{ and }x^2-3y+12=0$ may have a real common solution, are
$\textbf{(A) }4\text{ only}\qquad \textbf{(B) }-7,~4\qquad \textbf{(C) }0,~4\qquad \textbf{(D) }\text{no }y\qquad  \textbf{(E) }\text{all }y$
Because x2 + y2 + 16 = 0 has no real solutions, ∀ sets containing x2 + y2 + 16 = 0, no real solutions may exist.
∴ the solution is $\fbox{D}$

– TylerO_1.618