\(2x^2-2x+5x-5=0\\ \Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\end{matrix}\right.\\ Vậy......\)
\(2x^2+3x-5=0\\ \rightarrow2x^2+5x-2x-5=0\\ \rightarrow x\left(2x+5\right)-\left(2x+5\right)=0\\ \rightarrow\left(2x+5\right)\left(x-1\right)=0\\ \rightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\rightarrow\left[{}\begin{matrix}2x=-5\\x=1\end{matrix}\right.\rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(S=\left\{-\dfrac{5}{2};1\right\}\)