Find all equilibrium solutions (of the form x(t)=x0, a constant) of the second-order differential equation

x''+x'+x-6x^3=0

(Enter your answers in order of increasing value. Enter "na" for answer blanks you don't need.)
x=_____;x=_____;x=_____

I thought the answer would be 0, na, na but that is not correct
Any help would be great!

You have:

x'' + x' + x - 6x^3 = 0

We need to convert this second order equation to a system of first-order equations:

Let u = x', then u' = x''

Using this substitution, we can rewrite your second-order equation as a system of first-order equations:

dx/dt = u

du/dt = 6x^3 - u - x

Now solve for the equilibrium points of this system

dx/dt = u = 0
and
du/dt = 6x^3 - u - x = 0

The first of these equations means that an equilibrium solution must have u = 0 (not surprisingly). The second equation then reduces to:

6x^3 - x = 0

x(6x^2 - 1) = 0

x*(x*sqrt(6) + 1)(x*sqrt(6) - 1) = 0

So the equilibrium solutions are:

x = 0
x = +sqrt(1/6)
x = -sqrt(1/6)