#### Answer

$x=\left\{ -1-\sqrt{2},-1+\sqrt{2} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
t^2+2t-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}
t^2+2t=1
.\end{array}
In the equation above, $b=
2
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{2}{2} \right)^2
\\\\=
\left( 1 \right)^2
\\\\=
1
.\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}
t^2+2t+1=1+1
\\\\
t^2+2t+1=2
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(t+1)^2=2
.\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}
t+1=\pm\sqrt{2}
\\\\
t=-1\pm\sqrt{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
t=-1-\sqrt{2}
\\\\\text{OR}\\\\
t=-1+\sqrt{2}
.\end{array}
Hence, $
x=\left\{ -1-\sqrt{2},-1+\sqrt{2} \right\}
.$