Use interval notation to indicate the set of solutions to the inequality x² − 5x − 6 < 0.

Apply the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = 1$, $b = -5$, $c = -6$.

Compute the discriminant:

$$b^2 - 4ac = (-5)^2 - 4(1)(-6) = 25 + 24 = 49$$

Substitute:

$$x = \frac{5 \pm \sqrt{49}}{2} = \frac{5 \pm 7}{2}$$ $$x = \frac{5 + 7}{2} = \frac{12}{2} = 6, \quad x = \frac{5 - 7}{2} = \frac{-2}{2} = -1$$

The roots are $x = -1$ and $x = 6$.

• For $x = -2$ in $(-\infty, -1)$: $(-2 + 1)(-2 - 6) = (-1)(-8) = 8 > 0$
• For $x = 0$ in $(-1, 6)$: $(0 + 1)(0 - 6) = (1)(-6) = -6 < 0$
• For $x = 7$ in $(6, \infty)$: $(7 + 1)(7 - 6) = (8)(1) = 8 > 0$

The expression is negative or zero between the roots, and the parabola opens upwards (since $a > 0$).

Thus, $x^2 - 5x - 6 \leq 0$ for $-1 \leq x \leq 6$.

$[-1, 6]$.

Final answer: [-1,6]