16h5-25h3+9h = 0

Factor out common factor h

h(16h4-25h2+9) = 0

Factor the trinomial in the parentheses

h(16h2-9)(h2-1) = 0

Each of those binomial in parentheses are
the difference of squares, so factor them:

h(4h-3)(4h+3)(h-1)(h+1) = 0

By the zero-factor property set all those factors = 0:

h=0;  4h-3=0;  4h+3=0 ;   h-1=0;  h+1=0
        4h=3     4h=-3      h=1     h=-1
         h=      h=

So the solutions are 0, , , 1, and -1

Edwin