Find the product of the non-real roots of the equation \[(x^2-3)^2+5(x^2-3)+6=0.\]

$\begin{array}{@{\hspace{0em}}l@{\hspace{13.7em}}l@{\hspace{13.7em}}l} 

\textbf{(A) }0&\textbf{(B) }1&\textbf{(C) }-1\\\\ 

\textbf{(D) }2&\textbf{(E) }-2&\textbf{(F) }3\\\\

\textbf{(G) }-3&\textbf{(H) }4&\textbf{(I) }-4\\\\

\textbf{(J) }5&\textbf{(K) }-5&\textbf{(L) }6\\\\

\textbf{(M) }-6&\textbf{(N) }3+2i&\textbf{(O) }3-2i\\\\

\textbf{(P) }\dfrac{-3+i\sqrt3}2&\textbf{(Q) }8&\textbf{(R) }-8\\\\

\textbf{(S) }12&\textbf{(T) }-12&\textbf{(U) }42\\\\

\textbf{(V) }\text{Ying} & \textbf{(W) }2007 &\end{array}$
1. Let's start by simplifying the given equation:
   \[
   (x^2 - 3)^2 + 5(x^2 - 3) + 6 = 0
   \]
   Let \( y = x^2 - 3 \). Substituting \( y \) into the equation, we get:
   \[
   y^2 + 5y + 6 = 0
   \]

2. Next, we solve the quadratic equation \( y^2 + 5y + 6 = 0 \). We can factor this equation:
   \[
   y^2 + 5y + 6 = (y + 2)(y + 3) = 0
   \]
   Therefore, the solutions for \( y \) are:
   \[
   y = -2 \quad \text{and} \quad y = -3
   \]

3. Now, we substitute back \( y = x^2 - 3 \) to find the values of \( x \):
   - For \( y = -2 \):
     \[
     x^2 - 3 = -2 \implies x^2 = 1 \implies x = \pm 1
     \]
     These are real roots.
   - For \( y = -3 \):
     \[
     x^2 - 3 = -3 \implies x^2 = 0 \implies x = 0
     \]
     This is also a real root.

4. We need to find the product of the non-real roots. However, from the above steps, we see that all roots are real. Therefore, there are no non-real roots in this equation.

The final answer is \(\boxed{0}\)