\(\left( {x + 2} \right)\left( {x-3} \right) < 0\)
Решим неравенство методом интервалов:
\(\left( {x + 2} \right)\left( {x-3} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 2 = 0,}\\{x-3 = 0\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = 3.\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-2;3} \right).\)
Ответ: \(\left( {-2;\;3} \right).\)