Answer
$x=\left\{ 2-\sqrt{3},2+\sqrt{3} \right\}$
Work Step by Step
Using the properties of equality, the given equation, $
x^2-4x=-1
,$ is equivalent to
\begin{array}{l}\require{cancel}
x^2-4x+1=0
.\end{array}
The equation above has $a =1$,$ b = -4$, and $c=1$.
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions to the equation above are
\begin{array}{l}\require{cancel}
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}
\\\\
x=\dfrac{4\pm\sqrt{16-4}}{2}
\\\\
x=\dfrac{4\pm\sqrt{12}}{2}
\\\\
x=\dfrac{4\pm\sqrt{4\cdot3}}{2}
\\\\
x=\dfrac{4\pm\sqrt{(2)^2\cdot3}}{2}
\\\\
x=\dfrac{4\pm2\sqrt{3}}{2}
\\\\
x=\dfrac{2(2\pm\sqrt{3})}{2}
\\\\
x=\dfrac{\cancel{2}(2\pm\sqrt{3})}{\cancel{2}}
\\\\
x=2\pm\sqrt{3}
.\end{array}
Hence, the solutions are $
x=\left\{ 2-\sqrt{3},2+\sqrt{3} \right\}
.$
