Solve
-$\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A)$\{-24\}$ B)$\left\{\frac{10}{2}\right\}$ C)$\varnothing$ D)$\{-24,24\}$

Question

Solve 
-$\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A)$\{-24\}$ B)$\left\{\frac{10}{2}\right\}$ C)$\varnothing$ D)$\{-24,24\}$

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The Answer of Solve 
-$\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A)$\{-24\}$ B)$\left\{\frac{10}{2}\right\}$ C)$\varnothing$ D)$\{-24,24\}$

Solve
- $\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A) $\{-24\}$ B) $\left\{\frac{10}{2}\right\}$ C) $\varnothing$ D) $\{-24,24\}$

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Solve - log⁡6(x2+3x+19)=log⁡6(x2+x−29)\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)log6​(x2+3x+19)=log6​(x2+x−29) A) {−24}\{-24\}{−24} B) {102}\left\{\frac{10}{2}\right\}{210​} C) ∅\varnothing∅ D) {−24,24}\{-24,24\}{−24,24}

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Solve
- $\log _{6}\left(x^{2}+3 x+19\right)=\log _{6}\left(x^{2}+x-29\right)$ A) $\{-24\}$ B) $\left\{\frac{10}{2}\right\}$ C) $\varnothing$ D) $\{-24,24\}$

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