1. Find the roots

Given $x^2-x+2=0.$

Using the quadratic formula, $x=\frac{1\pm \sqrt{1-8}}{2}=\frac{1\pm \sqrt{-7}}{2}=\frac{1\pm i\sqrt7}{2}.$

Since $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$, we take $\alpha=\frac{1+i\sqrt7}{2},\qquad \beta=\frac{1-i\sqrt7}{2}.$

2. Use relations satisfied by the roots

Since $\alpha$ and $\beta$ are roots of $x^2-x+2=0$, each root satisfies $r^2-r+2=0 \implies r^2=r-2.$

So for $r=\alpha$, $\alpha^2=\alpha-2.$

Also, since $\beta$ is the conjugate of $\alpha$, $\beta^4=\overline{\alpha^4}.$ But an easier route is to express all powers using $r^2=r-2$.

3. Compute powers of $\alpha$

From $\alpha^2=\alpha-2,$ we get $\alpha^4=(\alpha^2)^2=(\alpha-2)^2=\alpha^2-4\alpha+4.$ Now substitute $\alpha^2=\alpha-2$: $\alpha^4=(\alpha-2)-4\alpha+4=-3\alpha+2.$

Next, $\alpha^6=\alpha^4\alpha^2=(-3\alpha+2)(\alpha-2).$ Expanding, $\alpha^6=-3\alpha^2+8\alpha-4.$ Substitute $\alpha^2=\alpha-2$: $\alpha^6=-3(\alpha-2)+8\alpha-4=5\alpha+2.$

4. Compute $\beta^4$

Similarly, $\beta^2=\beta-2,$ so $\beta^4=(\beta-2)^2=\beta^2-4\beta+4=(\beta-2)-4\beta+4=-3\beta+2.$

5. Evaluate the expression

We need $E=\alpha^6+\alpha^4+\beta^4-5\alpha^2.$ Substitute the values found: $E=(5\alpha+2)+(-3\alpha+2)+(-3\beta+2)-5(\alpha-2).$

Now simplify: $E=5\alpha+2-3\alpha+2-3\beta+2-5\alpha+10$ $E=(-3\alpha-3\beta)+16.$

Since $\alpha+\beta=1,$ we get $E=-3(1)+16=13.$

6. Final answer

$\boxed{13}$