Step 1: Factor the quadratic expression:

Step 2: The roots are $x = 1$ and $x = \frac{4}{3}$. Since it is $\le 0$, the solution lies between the roots:

Step 1: Factor the quadratic expression:

Step 2: The critical values are $x = -\frac{2}{3} \approx -0.666...$ and $x = 3$.

Step 3: For the inequality to be negative, $x$ must lie between the roots:

Step 1: Calculate the discriminant $D = b^2 - 4ac$:

Step 2: Since $D < 0$ and the leading coefficient $a = 3 > 0$, the quadratic expression is positive for all real $x$. Therefore, $3x^2 - 7x + 6$ can never be less than zero.

Step 1: Calculate the discriminant:

Step 2: Since $D < 0$ and $a = 1 > 0$, the parabola opens upwards and never crosses the x-axis. Thus, the expression is always positive for all real $x$.

Step 1: Factor the expression:

Step 2: The roots are $x = -1$ and $x = 15$. For the product to be positive, $x$ must lie outside the roots:

Step 1: Multiply the inequality by $-1$ (and flip the inequality sign):

Step 2: Factor:

Step 3: The roots are $x = -2$ and $x = 1$. The solution is:

Step 1: Split into double inequalities: $-5 < x^2 - 4x < 5$.

Part A: $x^2 - 4x > -5 \implies x^2 - 4x + 5 > 0$.

• Discriminant $D = 16 - 20 = -4 < 0$. Holds true for all real $x$.

Part B: $x^2 - 4x < 5 \implies x^2 - 4x - 5 < 0 \implies (x - 5)(x + 1) < 0$.

• Solution: $-1 < x < 5$.

Answer: (d) –1 < x < 5

Step 1: Rearrange to $\vert{}x^2 + x\vert{} < 5 \implies -5 < x^2 + x < 5$.

Part A: $x^2 + x + 5 > 0$ ($D = 1 - 20 < 0$, true for all real $x$).

Part B: $x^2 + x - 5 < 0$.

• Roots are $x = \frac{-1 \pm \sqrt{21}}{2}$.

• Solution is $\frac{-1 - \sqrt{21}}{2} < x < \frac{-1 + \sqrt{21}}{2}$.

Conclusion: None of the listed options reflect this boundary range.

Step 1: Double inequality: $-6 < x^2 - 5x < 6$.

Part A: $x^2 - 5x + 6 > 0 \implies (x - 2)(x - 3) > 0 \implies x < 2 \text{ or } x > 3$.

Part B: $x^2 - 5x - 6 < 0 \implies (x - 6)(x + 1) < 0 \implies -1 < x < 6$.

Intersection of Part A and B: $(-1 < x < 2) \cup (3 < x < 6)$

Step 1: Since absolute value is non-negative, $x > 0$.

Step 2: $-x < x^2 - 2x < x$.

• Left: $x^2 - x > 0 \implies x(x - 1) > 0$. Since $x > 0$, this requires $x > 1$.

• Right: $x^2 - 3x < 0 \implies x(x - 3) < 0$. Since $x > 0$, this requires $x < 3$.

Step 3: Combining yields $1 < x < 3$.