solve the differential equation $2y^{(5)}-7y^{(4)}+12y'''-8y''=0$. I cannot find the roots of the characteristic equation to get a solution.  I only know the basic way to solve these equations.  I factored out an $r^2$.
$2r^5-7r^4+12r^3-8r^2 = 0$
$r^2(2r^3-7r^2+12r-8) = 0$

Assuming $y^{\left(n\right)}\equiv\frac{\partial^{n}y}{\partial t^{n}}$.
Let $w=y^{\prime\prime}$ so that the equation becomes
$$
2w^{\prime\prime\prime}-7w^{\prime\prime}+12w^{\prime}-8w=0.
$$
The characteristic polynomial, as you pointed out, is
$$
2r^{3}-7r^{2}+12r-8
$$
with imaginary and real roots (solvable in $\mathbb{Q}$). As @Pieter21 points out, you can use Wolfram to get the roots; they are not "nice" numbers. Perhaps you can just use the approximate forms $1.2581$ and $1.210\pm 1.3867i$.
With a solution for $w$, you can derive a solution for $y$ since $y=\int\int w$.
