\(2x^2-3x-5=0\)
\(\Leftrightarrow2\left(x^2-\dfrac{3}{2}x-\dfrac{5}{2}\right)=0\)
\(\Leftrightarrow x^2-\dfrac{3}{2}x-\dfrac{5}{2}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\dfrac{49}{16}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right)^2=\dfrac{49}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}\\x-\dfrac{3}{4}=-\dfrac{7}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\end{matrix}\right.\)
