jmc

We start by completing the square: $$\begin{array}{rl}{x}^2+11x-5& =x^2+11x+{\left(\frac{11}{2}\right)}^2-{\left(\frac{11}{2}\right)}^2-5\\ & =x^2+11x+{\left(\frac{11}{2}\right)}^2-{\left(\frac{11}{2}\right)}^2-5\\ & ={\left(x+\frac{11}{2}\right)}^2-5-\frac{121}{4}.\end{array}$$Since the square of a real number is at least 0, we have ${\left(x+\frac{11}{2}\right)}^2\ge 0$, where ${\left(x+\frac{11}{2}\right)}^2=0$ only if $x=-\frac{11}{2}$. Therefore, the expression is minimized when $x=\boxed{-\frac{11}{2}}.$