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=>

\(10x^{-2}+x^{-1}-21=0\)
\(=> 10+x-21x^2=0\) by multiplying \(x^2\)
\(=> 21x^2-x-10=0\) by multiplying \((-1)\)
\(=> (3x+2)(7x-5)=0\)
\(=> x =\frac{-2}{3}\) or \(x = \frac{5}{7}\)

Therefore, the answer is A.
Answer: A
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A simpler hacky method to solve this using ACs:
For a quadratic equation ax^2+bx+c=0
---> Sum of roots = -b/a
---> Product of roots. = c/a

For the equation 21x^2−x−10=0, sum=(1/21) and product = (-10/21). This eliminates all options other than A
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