Solve the following equation and check for extraneous solutions. $\sqrt[4]{6 x^{2}-8}=-x$ Enter the solution(s) below. Round your answer to three decimal places. There are four solutions. They are: \[ \begin{array}{l} x_{1}= \\ x_{2}= \\ x_{3}= \\ x_{4}= \\ \end{array} \] There are two solutions. They are: \[ x_{1}=\quad x_{2}= \] There is one solution. It is: \[ x_{1}= \] There is no solution.

Step 1 ：The given equation is \(\sqrt[4]{6 x^{2}-8}=-x\).

Step 2 ：Square both sides to get rid of the fourth root, resulting in \(6x^2 - 8 = x^2\).

Step 3 ：Solve this quadratic equation to find the possible values of x.

Step 4 ：Check these solutions in the original equation to make sure they are not extraneous solutions.

Step 5 ：After solving, we found two potential solutions: -2 and approximately -1.414.

Step 6 ：However, only -2 was a valid solution. The other solution was extraneous.

Step 7 ：Final Answer: There is one solution. It is: \(x_{1}=\boxed{-2}\)