= Solution
The displayed equation has nominal orders $p=1,q=2$, with
$$
\Phi(z)=1-0.5z,\qquad
\Theta(z)=1-1.4z+0.45z^2=(1-0.5z)(1-0.9z).
$$
The AR root is $2$, and the MA roots are $2$ and $10/9$, all strictly outside the unit disc. Thus its stationary solution is causal and invertible. There is also <common factor cancellation in an ARMA model>: the convergent inverse of $1-0.5B$ cancels the shared factor, giving
$$
\boxed{x_t=w_t-0.9w_{t-1}.}
$$
Consequently \b[the displayed representation is ARMA(1,2), but its minimal order is ARMA(0,1)]. This also follows without formal operator cancellation: the difference between any solution and the right-hand side satisfies $d_t=0.5d_{t-1}$. A stationary finite-variance difference must have zero <variance>, since stationarity would give $\operatorname{Var}(d_t)=0.25\operatorname{Var}(d_t)$; its constant mean must also vanish. A transient chosen by an arbitrary initial condition is not the stationary process whose ACF is requested.
For this <moving-average model>, <independence> of the noise gives
$$
\gamma(0)=(1+0.9^2)\sigma_w^2=1.81\sigma_w^2,\qquad
\gamma(1)=\gamma(-1)=-0.9\sigma_w^2,
$$
while $\gamma(h)=0$ for $|h|\ge2$, since the two noise sets are disjoint. Therefore its <autocorrelation function> is
$$
\boxed{\rho(h)=\begin{cases}
1,&h=0,\\
-90/181,&|h|=1,\\
0,&|h|\ge2.
\end{cases}}
$$
The reduced moving-average inverse is $w_t=\sum_{j\ge0}0.9^jx_{t-j}$, whose coefficients are absolutely summable, confirming invertibility directly.
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