To solve the inequality $(4x + 3)(3x - 1) < 2x(x + 3)$, we need to first expand both sides and then move all terms to one side to set the inequality to zero. Let's work through it step by step:

1. Expand both sides:

   Left side:
   $(4x + 3)(3x - 1) = 12x^2 - 4x + 9x - 3 = 12x^2 + 5x - 3$

   Right side:
   $2x(x + 3) = 2x^2 + 6x$

2. Set the inequality to zero by moving all terms to the left side:

   $12x^2 + 5x - 3 - (2x^2 + 6x) < 0$

   Simplify the expression:

   $(12x^2 - 2x^2) + (5x - 6x) - 3 < 0$

   $10x^2 - x - 3 < 0$

3. Solve the quadratic inequality:

   We need to factor or use the quadratic formula to solve for $x$. The quadratic formula is given by:

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

   For our inequality $10x^2 - x - 3 = 0$, the coefficients are $a = 10$, $b = -1$, and $c = -3$.

   Calculate the discriminant ($b^2 - 4ac$):

   $(-1)^2 - 4(10)(-3) = 1 + 120 = 121$

   Now, solve for $x$:

   $x = \frac{-(-1) \pm \sqrt{121}}{2 \cdot 10}$

   $x = \frac{1 \pm 11}{20}$

   This gives us two solutions:

   $x_1 = \frac{1 + 11}{20} = \frac{12}{20} = \frac{3}{5}$

   $x_2 = \frac{1 - 11}{20} = \frac{-10}{20} = -\frac{1}{2}$

4. Determine the intervals for which the inequality is true:

   The roots divide the number line into three intervals: $(- \infty, -\frac{1}{2})$, $(- \frac{1}{2}, \frac{3}{5})$, and $(\frac{3}{5}, \infty)$.

   To find where $10x^2 - x - 3 < 0$, we need to test values from each interval:

   • For $x < -\frac{1}{2}$, choose $x = -1$:

     $10(-1)^2 - (-1) - 3 = 10 + 1 - 3 = 8 > 0$

   • For $-\frac{1}{2} < x < \frac{3}{5}$, choose $x = 0$:

     $10(0)^2 - (0) - 3 = -3 < 0$

   • For $x > \frac{3}{5}$, choose $x = 1$:

     $10(1)^2 - (1) - 3 = 10 - 1 - 3 = 6 > 0$

   The inequality $10x^2 - x - 3 < 0$ is satisfied in the interval $(- \frac{1}{2}, \frac{3}{5})$.

5. Conclusion:

   The solution to the inequality $(4x + 3)(3x - 1) < 2x(x + 3)$ is:

   $\boxed{-\frac{1}{2} < x < \frac{3}{5}}$

Would you like more details, or do you have any questions about this solution?

---

Here are some related questions to explore:

1. How do you solve quadratic inequalities using the sign test method?
2. What is the difference between solving equations and inequalities?
3. How do we know which intervals to choose when testing values for inequalities?
4. What are the key differences between linear and quadratic inequalities?
5. How can the quadratic formula help in finding the roots of a quadratic equation?

Tip: Always check the critical points and test intervals when solving inequalities to ensure the solution is accurate.