we can use the equation to factor
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a=6 , b=k and c=6
replacing
[tex]\begin{gathered} x=\frac{-k\pm\sqrt[]{k^2-4(6)(6)}}{2(6)} \\ \\ x=\frac{-k\pm\sqrt[]{k^2-4(6)(6)}}{2(6)} \\ \\ x=\frac{-k\pm\sqrt[]{k^2-144}}{12} \end{gathered}[/tex]to have a real solution we need the interior of the root to be greater than or equal to 0
so
[tex]\begin{gathered} k^2-144\ge0 \\ k^2\ge144 \\ k\ge\pm\sqrt[]{144} \\ k\le-12ork\ge12 \end{gathered}[/tex]The solution is option B
k can take any value in the interval K<-12 or in the interval k>12
