Why does this transfer function has a second zero I'm learning about $\mathcal Z$-transforms in DSP and I have a transfer function of the following form:

$$H(z)=\frac{2-3z^{-1}}{1-1.6z^{-1}+0.8z^{-2}}$$

When I calculate zeros and poles of this system by hand, 
I get these poles from the equation 

\begin{align}
\frac{z^2-1.6z+0.8}{z^2} &amp;= 0\\
p_1&amp;=0.8-0.4i\\
p_2&amp;=0.8+0.4i
\end{align}

And a single zero from the equation

\begin{align}
\frac{2z-3}{z}&amp;=0\\
z_1&amp;=1.5+0i
\end{align}

However, when I use Wolfram Alpha to compute the poles and zeros, it also lists a second zero positioned at origin. Or in MATLAB:



The question is, where does that second zero come from?
Maybe this way it is simpler to understand:

$$1-1.6z^{-1}+0.8z^{-2}=\frac{z^2-1.6z+0.8}{z^2}$$

Hence,

$$\frac{2-3z^{-1}}{1-1.6z^{-1}+0.8z^{-2}}=\frac{2-3z^{-1}}{\frac{z^2-1.6z+0.8}{z^2}}=\frac{z^2(2-3z^{-1})}{z^2-1.6z+0.8}$$

whose numerator is $2z^2-3z$.