#### Answer

$x=3\pm2\sqrt{5}$

#### Work Step by Step

$x^{2}-6x-11=0$
Let's take the independent term to the right side of the equation:
$x^{2}-6x=11$
Let's remember that, in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. In this equation, $b=-6$.
$x^{2}-6x+(\dfrac{-6}{2})^{2}=11+(\dfrac{-6}{2})^{2}$
$x^{2}-6x+9=11+9$
$x^{2}-6x+9=20$
We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes:
$(x-3)^{2}=20$
Take the square root of both terms:
$\sqrt{(x-3)^{2}}=\sqrt{20}$
$x-3=\pm\sqrt{20}$
$x=3\pm\sqrt{20}$
Since $\sqrt{20}=\sqrt{(4)(5)}=2\sqrt{5}$, we can also write the solution like this:
$x=3\pm2\sqrt{5}$