Answer
$\color{blue}{1 +i\sqrt{3},\ -1 -i\sqrt{3},\ -1 +i\sqrt{3},\ 1 - i\sqrt{3}}$
Work Step by Step
Rewrite the given equation as $(x^2)^2 + 4(x^2) +16=0$, let $u=x^2$ so that the given equation becomes $u^2+4u+16=0$, then use the quadratic equation to find solutions for $u$.
$\begin{align*}
u &= \frac{-4 \pm \sqrt{4^2-4(1)(16)}}{2(1)} \\
&= \frac{-4\pm \sqrt{16-64}}{2} \\
&= \frac{-4\pm \sqrt{-48}}{2} \\
&= \frac{-4\pm 4i\sqrt{3}}{2} \\
u &= -2 \pm 2i\sqrt{3} \qquad \text{(Eq.1)}
\end{align*}$
The four roots may now be obtained by using DeMoivre's Theorem to extract the square roots of each value for $u$ in Eq. 1.
$u = -2\pm 2i\sqrt{3} = x + iy \implies x=-2, y=\pm2\sqrt{3}$
$r = \sqrt{x^2+y^2} = \sqrt{(-2)^2 +(\pm2\sqrt{3})^2}= \sqrt{4+12} = \sqrt{16} = 4$
$\tan\theta = \pm y/x = \pm 2\sqrt{3}/2 = \pm \sqrt{3} \implies \theta = \underbrace{2\pi/3}_{x<0,\ y>0}, \underbrace{4\pi/3}_{x<0,\ y<0}$
$\begin{align*}
u &= r\,\text{cis}\,\theta \\
u &= 4\,\text{cis}\,(2\pi/3), \color{blue}{4\,\text{cis}\,(4\pi/3)} \\
z^2 &= 4\,\text{cis}\, (2\pi/3), \color{blue}{4\,\text{cis}\,(4\pi/3)} \\
z &= 2\,\text{cis}\,(\pi/3 + k\pi), \color{blue}{2\,\text{cis}\,(2\pi/3 + k\pi)},\ k=0,1 \\
&= 2\,\text{cis}\,(\pi/3), 2\,\text{cis}\,(4\pi/3), \color{blue}{2\,\text{cis}\,(2\pi/3), 4\,\text{cis}\,(5\pi/3)} \\
&= 2(\cos(\pi/3) + i\sin(\pi/3)), 2(\cos(4\pi/3) +i\sin(4\pi/3)), \\
& \qquad\qquad \color{blue}{2(\cos(2\pi/3) +i\sin(2\pi/3)), 2(\cos(5\pi/3) + i\sin(5\pi/3))} \\
&= 2(1/2 + i(\sqrt{3}/2)), 2(-1/2 - i(\sqrt{3}/2)), \\
& \qquad\qquad \color{blue}{2(-1/2 + i(\sqrt{3}/2)), 2(1/2 - i(\sqrt{3}/2))} \\
z &= 1 +i\sqrt{3},\ -1 -i\sqrt{3}, \ \color{blue}{-1 +i\sqrt{3},\ 1 - i\sqrt{3}}
\end{align*}$
