If x is real, then the value of ${\mathrm{x}}^2-6\mathrm{x}+13$ will not be less than

Let $\mathrm{y}={\mathrm{x}}^2-6\mathrm{x}+13\Rightarrow {\mathrm{x}}^2-6\mathrm{x}+13-\mathrm{y}=0$

Its discriminant $\mathrm{D}\ge 0\Rightarrow 36-4(13-\mathrm{y})\ge 0$

$\Rightarrow 36-52+4\mathrm{y}\ge 0\Rightarrow 4\mathrm{y}\ge 16\Rightarrow \mathrm{y}\ge 4$

Hence y is not less than 4.