For what values of $k$ is the inequality $x^2 - \left(k-3\right) x - k + 6 > 0$ valid for all real $x$?
Consider the quadratic equation $\;\;$ $x^2 - \left(k-3\right)x - k + 6 = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ will have real roots when its discriminant
$\Delta = \left[- \left(k - 3\right)\right]^2 - 4 \times 1 \times \left(6 - k\right) > 0$
i.e. $\;$ $k^2 - 6k + 9 - 24 + 4k > 0$
i.e. $\;$ $k^2 - 2k - 15 > 0$
i.e. $\;$ $\left(k + 3\right) \left(k - 5\right) > 0$
i.e. $\;$ $k + 3 > 0$ $\;$ and $\;$ $k - 5 > 0$ $\;\;$ OR $\;\;$ $k + 3 < 0$ $\;$ and $\;$ $k - 5 < 0$
i.e. $\;$ $k > -3$ $\;$ and $\;$ $k > 5$ $\;\;$ OR $\;\;$ $k < -3$ $\;$ and $\;$ $k < 5$
i.e. $\;$ $k > 5$ $\;\;$ OR $\;\;$ $k < -3$
i.e. $\;$ $5 < k < -3$
$\therefore \;$ The given inequality is valid for all real $x$ when $k \in \left(-3, 5\right)$