solution: Let us first define $y=x^{2}$. We can then plug this value into the inequality and add 4 to $-4$, $x^4+4x^2$, and 21, yielding $$0<y^{2}+4y+4<25.$$We can factor $y^2+4y+4$ to obtain $$0<(y+2)^{2}<25.$$Taking the square root gives us $0<|y+2|<5$, which in turn gives us two intervals for solutions of $y$: $-2<y<3$, or $-7<y<-2$.
  
  However, $y$ must be non-negative since $y=x^{2}$, so we have $0\leq y<3$. This means that $-\sqrt{3}< x<\sqrt{3}$ satisfies the original inequality. In interval notation, this is $\boxed{(-\sqrt{3}, \sqrt{3})}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  For what real values of $x$ is $-4<x^{4}+4x^{2}<21$ satisfied? Express your answer in interval notation.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: (-\sqrt{3}, \sqrt{3})
extra_info.question: For what real values of $x$ is $-4<x^{4}+4x^{2}<21$ satisfied? Express your answer in interval notation.
extra_info.split: test