Answer:
The output is given as
![y(n)=x[n]-2x[n-1]+3x[n-2]](https://img.qammunity.org/2021/formulas/mathematics/college/9bustn8ht7ytw90yyd04u3pb32vouexshw.png)
Explanation:
As the output is given as y(n) and the input is given as x(n) thus the
equation of the output is given as

Here the impulse response h(n) is given as
![h(n)=\delta[n]-2\delta[n-1]+3\delta[n-2]](https://img.qammunity.org/2021/formulas/mathematics/college/omd1qmsauk9l1qgx7gilyek2ob2y5t170k.png)
So the output is given as
![y(n)=h(n)*x(n)\\y(n)=(\delta[n]-2\delta[n-1]+3\delta[n-2])*x(n)](https://img.qammunity.org/2021/formulas/mathematics/college/ov16p01z713svq3p6aboe0heq3qqw5y092.png)
By distributive property,
![(ax1[n] + bx2[n]) * h[n] = ax1[n] * h[n] + bx2[n] * h[n]](https://img.qammunity.org/2021/formulas/mathematics/college/m7u65f9oq21zv6303vaci3eaq7sfhsbpf7.png)
![y(n)=h(n)*x(n)\\y(n)=\delta[n]*x(n)-2\delta[n-1]*x(n)+3\delta[n-2]*x(n)](https://img.qammunity.org/2021/formulas/mathematics/college/c558qjpn8m94u40k3ucocsv3d4raeh7e9k.png)
Now by the convolution properties with delta
![\delta[n- k] *x[n] = x[n-k]](https://img.qammunity.org/2021/formulas/mathematics/college/p5jg7x59xy0tw4df91jlboqr1v1blua1bh.png)
so
![y(n)=\delta[n]*x(n)-2\delta[n-1]*x(n)+3\delta[n-2]*x(n)\\y(n)=\delta[n-0]*x(n)-2\delta[n-1]*x(n)+3\delta[n-2]*x(n)\\y(n)=x[n]-2x[n-1]+3x[n-2]](https://img.qammunity.org/2021/formulas/mathematics/college/58vip3y35vxmqw8dgpnh0qjoqkc882z5le.png)
So the output is given as
![y(n)=x[n]-2x[n-1]+3x[n-2]](https://img.qammunity.org/2021/formulas/mathematics/college/9bustn8ht7ytw90yyd04u3pb32vouexshw.png)