Answer
$x=1+\sqrt{7}$ or $x=1-\sqrt{7}$
Work Step by Step
$ x^{2}-2x=6\qquad$...square half the coefficient of $x$ and add it to both sides to complete the square
$(\displaystyle \frac{-2}{2})=-1,\ (-1)^{2}=1$
$ x^{2}-2x+1=7\\qquad$...write $x^{2}-2x+1$ as a binomial squared.
$(x-1)^{2}=7\qquad$...use the principle of square roots:
For any real number $k$ and any algebraic expression $X$ :
$\text{If }X^{2}=k,\text{ then }X=\sqrt{k}\text{ or }X=-\sqrt{k}$
$ x-1=\pm\sqrt{7}\qquad$...add $1$ to both sides.
$ x=1\pm\sqrt{7}\qquad$... the symbol $\pm$ indicates two solutions.
$x=1+\sqrt{7}$ or $x=1-\sqrt{7}$
