This is done in the solution of exercise in order to make it possible to do inverse Laplace transform. Though I am not sure how is that done, so here it is:
$$\frac{s^2+3s+3}{2s^2+7s+7}=\frac{1}{2}-\frac{s+1}{2(2s^2+7s+7)}=\frac{1}{2}-\frac{0.5s+0.5}{2s^2+7s+7}=\frac{1}{2}-\frac{0.5s+0.5}{(s+1.75)^2+0.6614^2}=\frac{1}{2}-\frac{0.5(s+1.75)}{(s+1.75)^2+0.6614^2}+(0.5\cdot1.75-0.5)\cdot \frac{1}{0.6614}\cdot\frac{0.6614}{(s+1.75)^2+0.6614^2}$$
Here is now obviously possible to do inverse Laplace transform.
I can see that point is to get rid of $s^2$ from numerator by extracting what stands by $s^2$ in denominator, but then I don't understand why we have to subtract it from $\frac{1}{2}$.
Please if someone could explain it detailed, I would be very grateful.
Here is one way to think of it: if we multiply throughout by $\frac{2}{2} = 1$ we get
$$\frac{1}{2}\cdot\frac{2s^2 + 6s + 6}{2s^2 + 7s + 7}$$ $$= \frac{1}{2}\cdot\frac{(2s^2 + 7s + 7) - s - 1}{2s^2 + 7s + 7}$$ $$= \frac{1}{2}\cdot\left(1 + \frac{-s-1}{2s^2 + 7s + 7}\right)$$ $$= \frac{1}{2}\cdot\left(1 - \frac{s + 1}{2s^2 + 7s + 7}\right)$$ $$= \frac{1}{2} - \frac{s + 1}{2(2s^2 + 7s + 7)}$$
It is basically the divisiton of the numerator by the denominator, to bring out the quotient and remainder, to decompose the fraction. The above steps justifies the deduced results.