I don't have a graphing calculator and I don't know what the system of roots is.
But I can solve this anyway.
x^4 - 4x^3 - 6x^2 + 12 x = 0
x(x^3 - 4x^2 - 6x + 12) = 0
So x = 0 is a root and we have a cubic equation to solve.
In school the best way to solve cubic equations is to just try some small numbers; teachers seldom venture beyond 1, -1, 2 or -2 as roots. I guess a graphing calculator saves us the search.
Here we see x=-2 is a root; (-2)^3 - 4(-2)^2 - 6(-2) + 12 = -8 - 16 + 12 + 12 = 0.
So, continuing our factoring, x+2 is a factor, and we divide to get the other factor
x^2 - 6x + 6
x+2 | x^3 - 4x^2 - 6x + 12
x^3 + 2x^2
-6x^2 - 6x
- 6x^2 - 12 x
6x + 12
So fully factored we have
0 = x(x+2)( x^2 - 6x +6 )
There's a little shortcut I call the Shakespeare Quadratic Formula (2b or -2b) which works when the linear term is even:
[tex]x^2 - 2bx + c \textrm{ has zeros } x =b \pm \sqrt{b^2-c}[/tex]
Here that means we have two additional roots,
[tex]x = 3 \pm \sqrt{3^2-6} = 3 \pm \sqrt{3}[/tex]
Full set of roots:
[tex]x = 0 \textrm{ or } -2 \textrm{ or } 3 + \sqrt{3} \textrm{ or } 3 - \sqrt{3} [/tex]
