Answer
$x=\left\{ \dfrac{-7-\sqrt{53}}{2},\dfrac{-7+\sqrt{53}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
x^2+7x-1=0
,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
x^2+7x=1
.\end{array}
In the equation above, $b=
7
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{7}{2} \right)^2
\\\\=
\dfrac{49}{4}
.\end{array}
Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to
\begin{array}{l}\require{cancel}
x^2+7x+\dfrac{49}{4}=1+\dfrac{49}{4}
\\\\
x^2+7x+\dfrac{49}{4}=\dfrac{4}{4}+\dfrac{49}{4}
\\\\
x^2+7x+\dfrac{49}{4}=\dfrac{53}{4}
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( x+\dfrac{7}{2} \right)^2=\dfrac{53}{4}
.\end{array}
Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+\dfrac{7}{2}=\pm\sqrt{\dfrac{53}{4}}
\\\\
x+\dfrac{7}{2}=\pm\sqrt{\dfrac{1}{4}\cdot53}
\\\\
x+\dfrac{7}{2}=\pm\dfrac{1}{2}\sqrt{53}
\\\\
x+\dfrac{7}{2}=\pm\dfrac{\sqrt{53}}{2}
\\\\
x=-\dfrac{7}{2}\pm\dfrac{\sqrt{53}}{2}
\\\\
x=\dfrac{-7\pm\sqrt{53}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-7-\sqrt{53}}{2}
\\\\\text{OR}\\\\
x=\dfrac{-7+\sqrt{53}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{-7-\sqrt{53}}{2},\dfrac{-7+\sqrt{53}}{2} \right\}
.$