Find the inverse Laplace transform of (4s-6)/(s^2-2s+2).

Sun, I am afraid that this is not a "simple" inverse Laplace transform problem. it is not terribly difficult, but it requires quite a bit of calculations.....

The answer is e^{(1-i)}t[(2-i)+(2+i)e^{2it}]

and it requires that you use complex numbers. Note that the solutions of s^{2}-2s+2 = 0 are s = 1-i and s = 1+i. Hence you can decompose your assigned expresssion as

(4s-6)/[(s-(1-i))*(s-(1+i))]

and then you need to go from here using partial fractions, but in this case you will have to use complex numbers. If I am not mistaken, you get:

4s+6/(s^2-2s+2) = (2-5i)/(s-(1+i)) + (2+5i)/(s-(1-i))

and then you go from here.. But you need to be careful as you will be using complex numbers (technically you are closer to using Fourier transforms rather than Laplace transforms)...