Answer
$\color{blue}{\left\{-1-i, -1+i\right\}}$.
Work Step by Step
The complex zeros can be found using the quadratic formula, $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a, b, $ and $c$ are the coefficients of the terms of the trinomial $ax^2+bx+c$.
The given function involves a trinomial that has $a=1, b=2,$ and $c=2$. Substitute these values into the quadratic formula to obtain:
$x=\dfrac{-2\pm\sqrt{2^2-4(1)(2)}}{2(1)}
\\x=\dfrac{-2\pm\sqrt{4-8}}{2}
\\x=\dfrac{-2\pm\sqrt{-4}}{2}
\\x=\dfrac{-2\pm\sqrt{4(-1)}}{2}
\\x=\dfrac{-2\pm 2i}{2}
\\x=\dfrac{-2}{2} \pm \dfrac{2i}{2}
\\x=-1\pm i$
Thus, the complex zeros of the given function are: $\color{blue}{\left\{-1-i, -1+i\right\}}$.
