Inverse Laplace Transformation

I have the following equation: y"+6y'+8y=2. I am supposed to use laplace transformation and inverse laplace transformation to solve it. For la place transformation I got Y(s) = 2/s(s^2+6s+8). I'm confused how to find the inverse now. Any help???

Now, use partial fractions.
2/[s(s^2 + 6s + 8)]
= 2/[s(s + 2)(s + 4)] = A/s + B/(s + 2) + C/(s + 4) for some A, B, C.

Clearing denominators:
2 = A(s + 2)(s + 4) + Bs(s + 4) + Cs(s + 2)

s = 0 ==> 2 = 8A ==> A = 1/4
s = -2 ==> 2 = -4B ==> B = -1/2
s = -4 ==> 2 = 8C ==> C = 1/4.

So, 2/[s(s^2 + 6s + 8)] = (1/4) [1/s - 2/(s + 2) + 1/(s + 4)]

Inverting term by term yields
y = (1/4)[1 - 2e^(-2t) + e^(-4t)].

I hope this helps!

Partial fractions :)
The denominator facotrs to :

2/[(s)(s+2)(s+4) = a/s + b/(s+2) + c/(s+4).

a(s+2)(s+4) + b(s)(s+4) + c(s)(s+2) = as^2 + 6a*s + 8a + bs^2 + 4bs +cs^2 + 2cs. =2.
The term without 's' is "8a" which is therefore equal to 2. a = 1/4.

(a+b+c) [the coefficients of s^2] =0.
6a+4b+2c [the coefficients of s] = 0.

b+c = -1/4.
2b+2c= -1/2
4b+2c= -3/2.

-2b = 1
b=-1/2.
c=1/4.
= (1/4)/s + (-1/2)/(s+2) + (1/4)/(s+4). Inverse laplacing gets you: 1/4 -1/2*e^(-2t) +1/4*e^(-4t)