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64 6x 2 0

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64-6x^2>0

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Answer to a math question 64-6x^2>0

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$-6{x}^{2} > -64$

${x}^{2} < \frac{ 32 }{ 3 }$

$|x| < \sqrt{ \frac{ 32 }{ 3 } }$

$|x| < \frac{ \sqrt{ 32 } }{ \sqrt{ 3 } }$

$|x| < \frac{ {2}^{2}\sqrt{ 2 } }{ \sqrt{ 3 } }$

$|x| < \frac{ 4\sqrt{ 2 } }{ \sqrt{ 3 } }$

$|x| < \frac{ 4\sqrt{ 6 } }{ 3 }$

$\begin{array} { l }\begin{array} { l }x < \frac{ 4\sqrt{ 6 } }{ 3 },& x \geq 0\end{array},\\\begin{array} { l }-x < \frac{ 4\sqrt{ 6 } }{ 3 },& x < 0\end{array}\end{array}$

$\begin{array} { l }x \in \left[ 0, \frac{ 4\sqrt{ 6 } }{ 3 }\right\rangle,\\\begin{array} { l }-x < \frac{ 4\sqrt{ 6 } }{ 3 },& x < 0\end{array}\end{array}$

$\begin{array} { l }x \in \left[ 0, \frac{ 4\sqrt{ 6 } }{ 3 }\right\rangle,\\\begin{array} { l }x > -\frac{ 4\sqrt{ 6 } }{ 3 },& x < 0\end{array}\end{array}$

$\begin{array} { l }x \in \left[ 0, \frac{ 4\sqrt{ 6 } }{ 3 }\right\rangle,\\x \in \langle-\frac{ 4\sqrt{ 6 } }{ 3 }, 0\rangle\end{array}$

$\begin{align*} & x\in\langle-\frac{ 4\sqrt{ 6 } }{ 3 },\frac{ 4\sqrt{ 6 } }{ 3 }\rangle \\ & \end{align*}$

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