\[x^{2} - 12x + 39 > 0\]
\[\left( x^{2} - 12x + 36 \right) + 3 > 0\]
\[(x - 6)^{2} + 3 > 0\]
\[(x - 6)^{2} \geq 0\]
\[\Longrightarrow (x - 6)^{2} + 3 > 0.\]
\[Ответ:x - любое\ число.\]