Answer
$\{2,6\}$
Work Step by Step
Distribute the parentheses
$x^{2}-8x+12=0$
Factor the LHS
For $x^{2}+bx+c$ we search for two factors of c, such that their sum is b...
For c=12, the two factors are $-6$ and $-2$
$(x-6)(x-2)=0$
By the zero product principle, x may be $6$ or $2.$
Solution set: $\{2,6\}$
