#### Answer

$x=\left\{ 3-\sqrt{2},3+\sqrt{2} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
x^2-6x=-7
,$ express the equation in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
x^2-6x+7=0
.\end{array}
In the equation above, $a=
1
,$ $b=
-6
,$ and $c=
7
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(7)}}{2(1)}
\\\\
x=\dfrac{6\pm\sqrt{36-28}}{2}
\\\\
x=\dfrac{6\pm\sqrt{8}}{2}
.\end{array}
Extracting the perfect square factor of the radicand results to
\begin{array}{l}\require{cancel}
x=\dfrac{6\pm\sqrt{4\cdot2}}{2}
\\\\
x=\dfrac{6\pm\sqrt{(2)^2\cdot2}}{2}
\\\\
x=\dfrac{6\pm2\sqrt{2}}{2}
\\\\
x=\dfrac{2(3\pm\sqrt{2})}{2}
\\\\
x=\dfrac{\cancel2(3\pm\sqrt{2})}{\cancel2}
\\\\
x=3\pm\sqrt{2}
.\end{array}
The solutions are $
x=\left\{ 3-\sqrt{2},3+\sqrt{2} \right\}
.$