[tex]x^2-5x-1=0\\\\a=1;\ b=-5;\ c=-1\\\Delta=b^2-4ac\\\\\Delta=(-5)^2-4\cdot1\cdot(-1)=25+4=29 \ \textgreater \ 0\\\\therefore\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ \dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{5-\sqrt{29}}{2\cdot1}=\boxed{\dfrac{5-\sqrt{29}}{2}}\\\\x_2=\dfrac{5+\sqrt{29}}{2\cdot1}=\boxed{\dfrac{5+\sqrt{29}}{2}}[/tex]
