$x=\pm\sqrt3$ or $x=2$ or $x=-2$
Work Step by Step
$x^6-4x^4-9x^2+36=(x-2)(x+2)(x^2-3)(x^2+3)=0$ Thus $x^2+3=0$ (but this is impossible since $x^2$ is non-negative) or $x^2-3=0\\x^2=3\\x=\pm\sqrt3$ or $x-2=0\\x=2$ or $x+2=0\\x=-2$
You can help us out by revising, improving and updating this answer.Update this answer
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.