If the roots of the equation $x^2-8x+a^2-6a=0$  are real, then the  value of a will  be 

Given equation  is $8x^2-8x+a^2-6a=0 ... ( i )$

 If roots of Eq. (i) are real, then $D\ge 0$

  $\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-8{)}^2-4\times 1\left(a^2-6a\right)\ge 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}64-4\left(a^2-6a\right)\ge 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}64-4a^2+24a\ge 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4a^2-24a-64\le 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\left(a^2-6a-16\right)\le 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2-6a-16\le 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\because 4>0]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(a+2)(a-8)\le 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

  Equating all factors to zero, we get a= -2, 8 

$\Rightarrow a\in [-2,8]\text{ i.e. }-2\le a\le 8$