Answer
The solutions are $1\pm\sqrt{2}$
Work Step by Step
$x^{2}-2x=1$
Take the $1$ to the left side of the equation:
$x^{2}-2x-1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
In this case, $a=1$, $b=-2$ and $c=-1$.
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\dfrac{2\pm\sqrt{4+4}}{2}=...$
$...=\dfrac{2\pm\sqrt{8}}{2}=\dfrac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}$
