Question: How do I take inverse Laplace transform of $\frac{-2s+3}{s^2-2s+2}$?
I have checked my transform table and there is not a suitable case for this expression.

Answer: 
To perform the inverse Laplace transform you need to complete the square at the denominator, 
$$ s^2-2s+2=(s-1)^2+1$$
so you rewrite your expression as

$$\frac{-2s+3}{(s-1)^2+1}= -2 \frac{(s-1)}{(s-1)^2+1}+\frac{(3-2)}{(s-1)^2+1}$$

now these expressions are standard on tables "exponentially decaying
 sine/cosine wave. The inverse transform is then 
$$e^t(-2\cos t+\sin t).$$
You may want to try this (slighlty) different approach:

Let $F(s)$ be the function to be inverse-Laplace transformed. Then, $F(s)$ admits the following partial fraction decomposition:

$$F(s) = \frac{A\_1}{s-s\_1} + \frac{A\_2}{s-s\_2},$$ where $s\_1 = 1-i$, $s\_2 = \overline{s}\_1 = 1+i$, are the complex roots of the denominator, $A\_1 = \lim\_{s\to s\_1} (s-s\_1) F(s) = -1+i/2$ and $A\_2 = \overline{A}\_1 = \lim\_{s\to s\_2}(s-s\_2) F(s) = -1-i/2$. Now take inverse Laplace's transform on each term to get:

$$f(t) = \mathcal{L}^{-1}[F(s)] = A\_1 e^{s\_1 t} + A\_2 e^{s\_2 t},$$ since $A\_1$ and $A\_2$ are complex conjugates as well as $s\_1$ and $s\_2$, the result is real, which you can easily prove with the help of Euler's identity: $e^{ix} = \cos{x} + i \sin{x}$.

I hope this helps.

Cheers!