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

Step 1 ：The inequality is $\sqrt[3]{4x^2-5}>-1$. 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 $x\le -\sqrt{5}/2$ or $x\ge \sqrt{5}/2$. 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: $\boxed{{\displaystyle x\le -1.12}}$ or $\boxed{{\displaystyle x\ge 1.12}}$.