Question

Answer 1

\(2x^2-5x-3=0\Rightarrow\left(x-3\right)\cdot\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

Ta có: \(2x^2-5x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)