a) \(-8x^2+23x+3=0\)
\(\Leftrightarrow8x^2-23x-3=0\)
\(\Leftrightarrow8x^2+x-24x-3=0\)
\(\Leftrightarrow x\left(8x+1\right)-3\left(8x+1\right)=0\)
\(\Leftrightarrow\left(8x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}8x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=-1\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{8}\\x=3\end{cases}}}\)
Vậy \(x\in\left\{-\frac{1}{8};3\right\}\)