Solution:

Answer: $\frac{1-\sqrt{2}}{2}$. Complete the square by adding $1$ to each side. Then $(x+1)^{2}=1+i=e^{\frac{i \pi}{4}} \sqrt{2}$, so $x+1= \pm e^{\frac{i \pi}{8}} \sqrt[4]{2}$. The desired product is then
$$\left(-1+\cos \left(\frac{\pi}{8}\right) \sqrt[4]{2}\right)\left(-1-\cos \left(\frac{\pi}{8}\right) \sqrt[4]{2}\right)=1-\cos ^{2}\left(\frac{\pi}{8}\right) \sqrt{2}=1-\frac{\left(1+\cos \left(\frac{\pi}{4}\right)\right)}{2} \sqrt{2}=\frac{1-\sqrt{2}}{2}$$