Show, by completing the square, that x² - 6x + 11 is greater than zero for every real number x.

To complete the square in $x^2 + bx$, add $\left(\frac{b}{2}\right)^2$: then $x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$. Move the number term first, add the same amount to both sides, and take the square root with $\pm$. If the square term has a number in front, divide by it first.