How to inverse laplace this

20.83 /(s^2 + 101.71s + 171)

[2083/100)[1/[s^2 + (10171/100)s + 171]]

Complete the square in the denominator:

[2083/100)[1/[(s + 10171/200)^2 - 96609241/40000 ]]
[2083/100)[1/[(s + 10171/200)^2 - (9829/200)^2]]

Multiply and divide by 9829/4166:

[4166/9829][[(9829/200)/[(s + 10171/200)^2 - (9829/200)^2]]
[4166/9829][[(9829/200)/[(s - (-10171/200))^2 - (9829/200)^2]]

Now, we have the following from a table:

L^(-1){[b] / [(s - a)^2 - b^2]} = [e^(at)][sinh(bt)]

Here, a = -10171/200, b = 9829/200:

[4166/9829][e^(-10171t / 200)][sinh(9829t / 200)]

Done!