Find the x-coordinates of any stationary points of the equation y = x^3 - 2x + 4/x

y = x^3 - 2x +4/x, dy/dx = 3x^2 - 2 - 4/(x^2) = 0 at the stationary points, 3x^4 - 2x^2 - 4 = 0, substitute in u for x^2: 3u^2 - 2u - 4 = 0, use the quadratic formula: u = (-(-2) +- sqrt((-2)^2 - 43(-4)))/2*3 = (2 +- sqrt(52))/6 = (1 +- sqrt(13))/3 = x^2, because 1-sqrt(13) < 0, the only real solutions for x are from x^2 = (1+sqrt(13))/3, x = +- sqrt((1+sqrt(13))/3)