Let $p(x)=\sqrt{-x}$, and $q(x)=8x^2+10x-3$. The domain of $p(q(x))$ can be written in the form $a\le x \le b$. Find $b-a$.
Question:
Let $p(x)=\sqrt{-x}$, and $q(x)=8x^2+10x-3$. The domain of $p(q(x))$ can be written in the form $a\le x \le b$. Find $b-a$.
Answer:
Let's think step by step.

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Let $p(x)=\sqrt{-x}$, and $q(x)=8x^2+10x-3$. The domain of $p(q(x))$ can be written in the form $a\le x \le b$. Find $b-a$.
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