[tex]\left(\sqrt{z^{2}+8} \right)^{2}=(1-2z)^{2}\\\\z^{2}+8=4z^{2}-4z+1\\\\0=3z^{2}-4z-7\\\\0=(z+1)(3z-7)\\\\z=-1, z=\frac{7}{3}[/tex]
We know that if z=7/3, the right-hand side will be negative, which means it is an extraneous solution as square roots cannot be negative.
