Solve the following equation for \( z \) \[ \sqrt{3 z^{2}+16}=\sqrt{2 z^{2}-8 z} \]

To solve the equation \[ \sqrt{3 z^{2}+16}=\sqrt{2 z^{2}-8 z}, \] we start by squaring both sides to eliminate the square roots. This gives us: \[ 3z^2 + 16 = 2z^2 - 8z. \] Now, we can rearrange the equation by bringing all terms to one side: \[ 3z^2 - 2z^2 + 8z + 16 = 0, \] which simplifies to \[ z^2 + 8z + 16 = 0. \] Next, we can factor the quadratic equation: \[ (z + 4)(z + 4) = 0, \] or equivalently, \[ (z + 4)^2 = 0. \] This implies that \[ z + 4 = 0 \quad \Rightarrow \quad z = -4. \] To ensure that this solution is valid, we substitute \( z = -4 \) back into the original equation: The left side becomes: \[ \sqrt{3(-4)^2 + 16} = \sqrt{3 \cdot 16 + 16} = \sqrt{48 + 16} = \sqrt{64} = 8. \] The right side becomes: \[ \sqrt{2(-4)^2 - 8(-4)} = \sqrt{2 \cdot 16 + 32} = \sqrt{32 + 32} = \sqrt{64} = 8. \] Since both sides equal 8, the solution \( z = -4 \) is valid. Thus, the solution is \[ \boxed{-4}. \]