This is an inhomogeneous linear equation, so the general solution may be expressed as a sum of the solution to the complementary equation (with zero on the right) and a particular (whatever) solution to the original equation.
Homogeneous part (\(y'' + 4 y' + 4 y = 0\)):
-- determine and factor characteristic polynomial: \(p(s) = s^2 + 4 s + 4 = (s + 2)^2\);
-- for a real root \(r\), a constant times \(e^{r t}\) is a solution, for a double root the same times \(t\) is also an (independent) solution;
-- the equation is of second order, so all complementary solutions are of the form: \(y_h = c_1 e^{-2t} + c_2 t e^{-2t}\) where \(c_1\) and \(c_2\) are some constants.
Inhomogeneous part -- we need just a single solution:
-- input is something times exponential, so lets guess that a solution is some unknown function times the same exponential: \(y_p = u(t) e^{-2t}\);
-- substitute the guess on the left: \(y_p'' + 4 y_p' + 4 y_p = (u'' - 4 u' + 4 u) e^{-2t} + 4 (u' - 2 u) e^{-2t} + 4 u e^{-2t} = u'' e^{-2t}\);
-- try to match it with the right: \(u'' e^{-2t} = t^2 e^{-2t}\), exponential is never equal to zero, so we can divide by it: \(u'' = t^2\);
-- and integrate twice to get \(u = \frac{t^4}{12}\) (we may ignore constants of integration because we need just a single solution);
-- \(y_p = \frac{t^4}{12} e^{-2t}\) is one function that solves your equation.
The general solution is sum of the two parts:
\(y = y_h + y_p = c_1 e^{-2t} + c_2 t e^{-2t} + \frac{t^4}{12} e^{-2t}\)
[exactly as Woodsmack has given above]
[In the second part, we could have guessed a polynomial (of degree 4 in this case) instead of using a general function, but this would result in even more calculation. In both cases, the session on undetermined coefficients describes a handy method of organizing those.]
This is still more a sketch than a full solution, but maybe it will help you understand where do things come from. If you'd like more details then maybe try Wolfram's Alpha -- it has a nice (and free) step-by-step solver.