#### Answer

$x=\pm 1$
$x=\pm \sqrt{3}$

#### Work Step by Step

$x^{4}-4x^{2}+3=0$
First, we substitute $y=x^{2}$ into the equation:
$y^{2}-4y+3=0$
Next, we factor and solve:
$(y-1)(y-3)=0$
$y-1=0$ or $y-3=0$
$y=1$ or $y=3$
Finally, we convert the $y$ solutions to $x$ solutions:
$y=x^2$
$x=\pm\sqrt{y}$
If $y=1$:
$x=\pm \sqrt{1}=\pm 1$
If $y=3$:
$x=\pm \sqrt{3}$