How to apply discrete filters to a signal
Another way to simply get your result for this kind of a problem (where $h[n]$ is very short) is to use the following method :

Let your output of the discrete convolution sum be $y[n]$ :
$$ y[n] = x[n] \star h[n] $$

Then by expanding $h[n]$ into impulses, the convolution will be distributed over addition ( using the highpass filter  {-1,2,-1}; k = -1,0,1;  to demonstrate ) :

$$ y[n] = x[n] \star \{ -\delta[n+1] + 2 \delta[n] - \delta[n-1] \} $$

$$ y[n] = - x[n+1] + 2 x[n] -x[n-1] $$

By simple argumentation of convolution nonzero ranges, it can be seen that $y[n]$ starts at the index $n=-1$ hence the first sample of the output is:

$$ y[-1] = -x[0] + 2 x[-1] - x[-2] = -x[0] = 20 $$

Note that $x[n]=0$ for $n&lt;0$; The second sample of $y[n]$ will be $y[0]$ which is:

$$ y[0] = -x[1] + 2 x[0] - x[-1] = 50 + -40 = 10 $$

and so on... You can apply the procedure for your other filter and other output samples