$4x^2 - 8x - 5 = 0$
$=> 4x^2 + 2x - 10x - 5 = 0$
$=> 2x(2x+1) - 5(2x+1) = 0$
$=> (2x-5)(2x+1) = 0$
=> \(\left[ \begin{array}{l}2x-5=0\\2x+1=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}2x=5\\2x=-1\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{array} \right.\)
Vậy $x $∈ { $\dfrac{5}{2}; \dfrac{-1}{2}$}
