Finding the output time response of a transfer function with a step response
You can do this in several ways but I like to use the Heaviside expansion formula more than the partial fraction decomposition. If you re-arrange your transfer function with a unitless denominator then multiplied by the step function \$\frac{1}{s}\$, you have \$D(s)=s(\frac{s^2}{144}+\frac{6}{144}s+1)\$. The forced value is the dc gain multiplied by the 1-V step and it is 20/144 or 139 mV.
The roots for this expression are \$s_1\$, \$s_2\$ and \$s_3\$. Now, find the derivative of \$D(s)\$ with respect to \$s\$ (call it \$D'(s)\$ and determine the value of \$D'(s_1)\$, \$D'(s_2)\$ and \$D'(s_3)\$. This is it, you have your time-domain expression if you apply the formula as shown below:

The next step is to re-arrange the complex exponential to unveil the sine and cosine terms but this is the extra mile I leave to the student legs. I like this method as it yields insight on how the time-domain response is directly affected by the natural time constants in the denominator. You also clearly see how the zeroes (the numerator roots) if any, affect the coefficients only while the decays are solely fixed by the poles.