How to find inverse laplace transform $$
F(s) = \dfrac{6s+9}{s^2-10s+29}
$$
How do you solve the inverse Laplace transform of this above equation?

$$
\underbrace{s^2-10s+29 = (s-5)^2+2^2}_{\text{completing the square}} = t^2+2^2
$$
So
$$
\frac{6s+9}{s^2-10s+29} = \frac{6(s-5)+39}{(s-5)^2+2^2} = \frac{6t+39}{t^2+2^2} = 6\frac{t}{t^2+2^2} + \frac{39}{2}\cdot\frac{2}{t^2+2^2}
$$
Now look at each term separately.
