We determine all zeros of the function $y(x)=(x-1)(x-2)(x-3)+6$.
$$\begin{align}
y(x)=0 & \Leftrightarrow (x-1)(x-2)(x-3)+6=0\\
& \Leftrightarrow (x^2-3x+2)(x-3)+6=0\\
& \Leftrightarrow x^3-6x^2+11x-6+6=0\\
& \Leftrightarrow x^3-6x^2+11x=0\\
& \Leftrightarrow x(x^2-6x+11)=0\\
& \Leftrightarrow x(x^2-6x+11)=0\\
& \Leftrightarrow x=0 \mbox{ or } x^2-6x+11=0
\end{align}$$
We determine the zeros of $x^2-6x+11$: $D=(-6)^2-4\cdot 1 \cdot 11=-8$. Hence, this part has no zeros.
Consequently, the only zero of $y(x)$ is $x=0$.
$$\begin{align}
y(x)=0 & \Leftrightarrow (x-1)(x-2)(x-3)+6=0\\
& \Leftrightarrow (x^2-3x+2)(x-3)+6=0\\
& \Leftrightarrow x^3-6x^2+11x-6+6=0\\
& \Leftrightarrow x^3-6x^2+11x=0\\
& \Leftrightarrow x(x^2-6x+11)=0\\
& \Leftrightarrow x(x^2-6x+11)=0\\
& \Leftrightarrow x=0 \mbox{ or } x^2-6x+11=0
\end{align}$$
We determine the zeros of $x^2-6x+11$: $D=(-6)^2-4\cdot 1 \cdot 11=-8$. Hence, this part has no zeros.
Consequently, the only zero of $y(x)$ is $x=0$.