If the roots of the equation ${\mathrm{x}}^2-8\mathrm{x}+({\mathrm{a}}^2-6\mathrm{a})=0$ are real, then

Roots of ${\mathrm{x}}^2-8\mathrm{x}+({\mathrm{a}}^2-6\mathrm{a})=0$ are real. So $\mathrm{D}\ge 0$

$$\begin{gathered} \Rightarrow 64-4({\mathrm{a}}^2-6\mathrm{a})\ge 0\Rightarrow 16-{\mathrm{a}}^2+6\mathrm{a}\ge 0 \\ \Rightarrow {\mathrm{a}}^2-6\mathrm{a}-16\le 0\Rightarrow (\mathrm{a}-8)(\mathrm{a}+2)\le 0 \end{gathered}$$

Now we have two cases:

Case I: $(\mathrm{a}-8)\le 0 \mathrm{and} (\mathrm{a}+2)\ge 0$

$\Rightarrow \mathrm{a}\le 8 \mathrm{and} \mathrm{a}\le -2$

Case II: $(\mathrm{a}-8)\ge 0 \mathrm{and} (\mathrm{a}+2)\le 0$

$\Rightarrow \mathrm{a}\ge 8 \mathrm{and} \mathrm{a}\le -2$ but it is impossible

Therefore, we get $-2\le \mathrm{a}\le 8$