Find the range of values of $\displaystyle x $for which $\displaystyle (1+x)(6-x) < = -8$
Quadratic Inequalities are ALWAYS easiest to solve by completing the square...
$\displaystyle \displaystyle (1 + x)(6 - x) \leq -8$
$\displaystyle \displaystyle 6 - x + 6x - x^2 \leq -8$
$\displaystyle \displaystyle x^2 - 5x - 6 \geq 8$
$\displaystyle \displaystyle x^2 - 5x \geq 14$
$\displaystyle \displaystyle x^2 - 5x + \left(-\frac{5}{2}\right)^2 \geq 14 + \left(-\frac{5}{2}\right)^2$
$\displaystyle \displaystyle \left(x - \frac{5}{2}\right)^2 \geq \frac{56}{4} + \frac{25}{4}$
$\displaystyle \displaystyle \left(x - \frac{5}{2}\right)^2 \geq \frac{81}{4}$.
You should be able to go from here. Also, note that "$\displaystyle \displaystyle \sqrt{X^2} = |X|$" and "$\displaystyle \displaystyle |X| \geq a$" is equivalent to "$\displaystyle \displaystyle X \leq -a$ and $\displaystyle \displaystyle X \geq a$".