hey guys,
voidflux.

Since the original equation is a quartic there will be 4 roots or values of x. You have already found 2 but these are real and imply that the graph of the function crosses the x axis at these points. The other 2 appear to be complex as there is a negative quantity under the square root sogn.
This is where what NotLKH has said comes in. By using the imaginary number i such that i^2 = -1 we can find the other roots(imaginary) so there is nothing worng with what you have done so far.
solving x^2 + x + 1 = 0 we get x = [-1 +/- sqrt(-3)] /2

now sqrt(-3) can be written as sqrt(3)i. therefor the four roots of the equation are:
x = 0 ,1, [-1 + sqrt(3)i]/2, [-1 - sqrt(3)i]/2