The graphs of $y=x^4$ and $y=5x^2-6+4x^2$ intersect at four points with $x$-coordinates $\pm \sqrt{m}$ and $\pm \sqrt{n}$, where $m > n$. What is $m-n$?
Setting the two equations equal and rearranging the terms gives $x^4 = 5x^2 + 4x^2 - 6$. Simplifying gives $x^4 - 9x^2 + 6 = 0$. This equation can be factored as $(x^2-3)(x^2 - 2)$. The four $x$-coordinates where the graphs of $y = x^4$ and $y= 5x^2 + 4x^2 - 6$ intersect are $\pm \sqrt{3}$ and $\pm \sqrt{2}$, so $m-n = \sqrt{3}^2 - \sqrt{2}^2 = 3-2=\boxed{1}$.
To find the $x$-coordinates of the points where the graphs intersect, we need to set the two equations equal to each other and solve for $x$:
$x^4 = 5x^2-6+4x^2$
Combining like terms:
$x^4 - 9x^2 + 6 = 0$
Now, let's factor this equation:
$(x^2 - 3)(x^2 - 2) = 0$
Setting each factor equal to zero:
$x^2 - 3 = 0$ or $x^2 - 2 = 0$
Solving for $x$ in the first equation:
$x = \pm \sqrt{3}$
And in the second equation:
$x = \pm \sqrt{2}$
Since $m$ is greater than $n$, we have:
$m = 3$ and $n = 2$
Therefore, $m - n = 3 - 2 = \boxed{1}$.