Answer
{$\dfrac{7}{2}\pm \dfrac{\sqrt{53}}{2}$}
Work Step by Step
Given: $x^2-7x-1=0$
or, $x^2-7x=1$
Add $(\dfrac{7}{2})^2$ on both sides to make a perfect square.
Thus,
$x^2-7x+(\dfrac{7}{2})^2=1+(\dfrac{7}{2})^2$
$x^2-2(x)(\dfrac{7}{2})+\dfrac{49}{4}=\dfrac{53}{4}$
or, $(x-\dfrac{7}{2})^2=\dfrac{53}{4}$
Apply Square root property.
$x=\dfrac{7}{2}\pm \dfrac{\sqrt{53}}{2}$
Hence,we have $x=${$\dfrac{7}{2}\pm \dfrac{\sqrt{53}}{2}$}
