The equation is: $x^4 + 4x^2 + 16 = 0$
I tried solving it by substitution and then using the quadratic formula: $x^2 = a$
$a^2 + 4a + 16 = 0$
using the quadratic formula I got $a = -2 \pm 2i\sqrt{3}$,
so $x = \pm \sqrt{-2 \pm 2i\sqrt{3}}$
But, the actual solution does not use substitution and has different results: $x^4 + 4x^2 + 16 = (x^2 + 2x + 4)(x^2 - 2x + 4)$
$$ x = \frac{-2 \pm \sqrt{4-4(4)(1)}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = - 1 \pm i\sqrt{3}$$
and
$$ x = \frac{2 \pm \sqrt{4-4(4)(1)}}{2} = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3}$$
So, the answer is $\pm 1 \pm i \sqrt{3}$
The question is, why was my approach wrong?