2) (x²+3)²-14(x²+3)+24=0;

Let z = x²+3. The equation becomes z² - 14z + 24 = 0. Factoring gives (z-2)(z-12) = 0. So, z=2 or z=12. Substituting back, x²+3=2 gives x²=-1, so x=±i. And x²+3=12 gives x²=9, so x=±3. The solutions are x = ±i, ±3.