Let $$\alpha, \beta$$ be the roots of the equation $$x^2 - x + 2 = 0$$ with $$\text{Im}(\alpha) > \text{Im}(\beta)$$. Then $$\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$$ is equal to _______

$$x^2 - x + 2 = 0$$: $$\alpha + \beta = 1, \alpha\beta = 2$$.

$$\alpha^2 + \beta^2 = 1 - 4 = -3$$. $$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = 9-8=1$$.

$$\alpha^6 = (\alpha^2)^3$$. $$\alpha^2 = \alpha - 2$$. $$\alpha^3 = \alpha^2 - 2\alpha = -\alpha-2$$.

$$\alpha^6 = (\alpha^3)^2 = (-\alpha-2)^2 = \alpha^2+4\alpha+4 = \alpha-2+4\alpha+4 = 5\alpha+2$$.

$$\alpha^6+\alpha^4+\beta^4-5\alpha^2 = 5\alpha+2+1-5(\alpha-2) = 5\alpha+3-5\alpha+10 = 13$$.

The answer is $$\boxed{13}$$.