Ans: D
$\begin{array}{rcl}
x^2 + 6x +k & = & 3 \\
x^2 + 6x + (k-3) & = & 0
\end{array}$
$\begin{array}{rcl}
x^2 + 6x +k & = & 3 \\
x^2 + 6x + (k-3) & = & 0
\end{array}$
Since the quadratic equation has no real root, then $\Delta <0$. Hence, we have
$\begin{array}{rcl}
(6)^2 – 4 (1) (k-3) & < & 0 \\
36 -4k + 12 & < & 0 \\
-4k & < & -48 \\
k & > & 12
\end{array}$
