Solve the following inequality.
\[
\sqrt[3]{4 x^{2}-5}> -1
\]
Round your answers to two decimal places.
$x< $ or $x> $
\[
< x<
\]

Step 1 ：The inequality is \(\sqrt[3]{4 x^{2}-5}>-1\). To solve this inequality, we need to first remove the cube root by cubing both sides of the inequality. This gives us \(4x^2 - 5 > -1\).

Step 2 ：We then isolate \(x^2\) by adding 5 to both sides, which gives us \(4x^2 > 4\).

Step 3 ：Finally, we divide both sides by 4 to solve for \(x^2\), which gives us \(x^2 > 1\). Taking the square root of both sides gives us \(x > 1\) or \(x < -1\).

Step 4 ：The solution to the inequality is \(x \leq -\sqrt{5}/2\) or \(x \geq \sqrt{5}/2\). However, the question asks for the solution to be rounded to two decimal places. Therefore, we need to evaluate \(\sqrt{5}/2\) and \(-\sqrt{5}/2\) and round the results to two decimal places.

Step 5 ：Final Answer: \(\boxed{x \leq -1.12}\) or \(\boxed{x \geq 1.12}\).