It is easy to note $h(1)=h(2)=0$.
It is evident from desmos that these are the only 2 solutions. I came across this function while trying to verify parallelogram law:
$2 \Vert f \Vert^2 + 2\Vert g \Vert^2 = \Vert f + g \Vert^2 + \Vert f - g \Vert^2$ where $\Vert F \Vert=(\int_0^1|F|^p)^{1/p}$ with $f=x$ and $g=1$
Taking the derivative of $h(x)$ makes everything very messy to deal with.
How do I show there is only 2 root to $h(x)$ for $x>0$?