If $\log_{10} (x^2-3x+6)=1$, the value of $x$ is:
$\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}\ 3\text{ or }-1 \qquad \textbf{(D)}\ 4\text{ or }-1\\ \textbf{(E)}\ \text{none of these}$
We know that $x^2-3x+6=10^1$, after expanding the logarithm. Taking the 10 to the other side, $x^2-3x-4=0$. Factoring, we get $(x-4)(x+1)=0$, so $x=\boxed{4,-1\Rightarrow \text{(D)}}$.