#### Answer

The solutions are $x=\pm1$ and $x=\pm\dfrac{4}{3}$

#### Work Step by Step

$9x^{4}=25x^{2}-16$
Take all terms to the left side of the equation:
$9x^{4}-25x^{2}+16=0$
Let $u$ be equal to $x^{2}$
If $u=x^{2}$, then $u^{2}=x^{4}$
Rewrite the equation using the new variable $u$:
$9u^{2}-25u+16=0$
Solve by factoring:
$(u-1)(9u-16)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-1=0$
$u=1$
$9u-16=0$
$9u=16$
$u=\dfrac{16}{9}$
Substitute $u$ back to $x^{2}$ and solve for $x$:
$x^{2}=1$
$\sqrt{x^{2}}=\pm\sqrt{1}$
$x=\pm1$
$x^{2}=\dfrac{16}{9}$
$\sqrt{x^{2}}=\pm\sqrt{\dfrac{16}{9}}$
$x=\pm\dfrac{4}{3}$
The solutions are $x=\pm1$ and $x=\pm\dfrac{4}{3}$