Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 596: 111

$x=\pm \sqrt 5,\pm \sqrt 3$

Need to solve for $y$. We have $x^4-8x^2+15=0$ or, $x^4-3x^2-5x^2+15=0$ After simplifications, we have, $(x^2-5)(x^2-3)=0$ Now, $(x^2-5)=0$ and $(x^2-3)=0$ or, $x^2=5$ and $x^2=3$ Hence, $x=\pm \sqrt 5,\pm \sqrt 3$