Ta có: \(x^2+5x=3\)
\(\Leftrightarrow x^2+5x-3=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{37}{4}=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{37}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{2}=-\frac{\sqrt{37}}{2}\\x+\frac{5}{2}=\frac{\sqrt{37}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\sqrt{37}-5}{2}\\x=\frac{\sqrt{37}-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-\sqrt{37}-5}{2};\frac{\sqrt{37}-5}{2}\right\}\)