Answer
$x=\dfrac{7}{2}\pm\dfrac{\sqrt{53}}{2}$
Work Step by Step
Using the completing the square method, then,
\begin{array}{l}
x^2-7x-1=0\\
x^2-7x=1\\
x^2-7x+\left( \dfrac{-7}{2} \right)^2=1+\left( \dfrac{-7}{2}\right)^2\\
x^2-7x+\dfrac{49}{4}=1+\dfrac{49}{4}\\
\left( x-\dfrac{7}{2} \right)^2=\dfrac{53}{4}\\
x-\dfrac{7}{2}=\pm\sqrt{\dfrac{53}{4}}\\
x=\dfrac{7}{2}\pm\dfrac{\sqrt{53}}{2}
.\end{array}
