Question

\(x^2-3x-5=0\)

Answer 1

\(x=\left\{\dfrac{3+\sqrt{29}}{2};\dfrac{3-\sqrt{29}}{2}\right\}\)

Answer 2

\(x^2-3x-5=0\\ \Delta=9+20=29\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{29}}{2}\\x=\dfrac{3+\sqrt{29}}{2}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{3+\sqrt{29}}{2};\dfrac{3-\sqrt{29}}{2}\right\}\)