Answer
$\left\{x\ | \ x \lt -5 \right\}$ or $\left(-\infty,-5\right)$
Work Step by Step
$(x+2)(x-3) \gt (x-1)(x+1)$
$x^{2}-x-6 \gt x^{2}-1 \qquad$ ... add $6-x^{2}$
$-x \gt 5\qquad$ ... multiply with $-1 $ (negative)
... the inequality changes direction ...
$x \lt -5$
Solution set: $\left\{x\ | x \lt -5 \right\}$ or $\left(-\infty,-5\right)$
