Question 180326
{{{9x^4-25x^2+16=0}}}
{{{(3x^2)^2-25x^2+16=0}}}
{{{3u^2-25u+16=0}}}
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Sub u for x^2
{{{9u^2 - 25u + 16 = 0}}}  You used 3u for one term, and u for the other.
You can factor"
(9u-16)*(u-1) = 0
u = 1, u = 16/9
Then
x = +1, -1
x = +4/3, -4/3
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Another approach is to "eyeball" the 9, -25, 16 and spot that +1 and -1 are solutions, factor those out, then you have a quadratic.