Find the solution of $y'''+y''-y'-4y=2-2x$, $y(1)=\frac{1}{2}, y'(1)=y''(1)=0$ by using Laplace transformation

Question

I want to find the solution of $y'''+y''-y'-4y=2-2x\ $, $y(1)=\frac{1}{2},\ y'(1)=y''(1)=0$ by using Laplace transformation. When I try to solve this exercise by appling Laplace transformation I get that $$L\{y(t)\}=\frac{1}{2}\frac{s^2+s-1}{s(s^2+s-1)-4}-\frac{2}{s(s^3+s^2-s-4)}-\frac{2}{s^2(s^3+s^2-s-4)} .$$ The problem is that $s^3+s^2-s-4 $ has not only real roots but complex roots, too, and of course I can not find them analitic. Does anyone has any ideas what's the next step? Thanks!!