We start with the equation:
\[ (6x^2)(2x + 3) = 4(2x + 3) + 5x(2x + 3) \]
First, notice that both sides of the equation share the common factor \( (2x + 3) \). We can factor this out:
\[ (2x + 3)(6x^2) = (2x + 3)(4 + 5x) \]
Assuming \( 2x + 3 \neq 0 \), we can divide both sides by \( 2x + 3 \):
\[ 6x^2 = 4 + 5x \]
Now, we will rearrange the equation to set it to 0:
\[ 6x^2 - 5x - 4 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula, where \( a = 6 \), \( b = -5 \), and \( c = -4 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant:
\[ b^2 - 4ac = (-5)^2 - 4(6)(-4) \] \[ = 25 + 96 \] \[ = 121 \]
Now, we can apply the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{121}}{2(6)} \] \[ = \frac{5 \pm 11}{12} \]
Calculating the two possible solutions:
\[ x = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3} \]
\[ x = \frac{5 - 11}{12} = \frac{-6}{12} = -\frac{1}{2} \]
Thus, the solutions to the equation are:
\[ \boxed{\left{ \frac{4}{3}, -\frac{1}{2} \right}} \]
Finally, we should check if \( 2x + 3 = 0 \) gives any additional solutions. Solving \( 2x + 3 = 0 \) gives:
\[ 2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2} \]
Thus, we have an additional solution:
\[ x = -\frac{3}{2} \]
The final set of solutions is therefore:
\[ \boxed{\left{ \frac{4}{3}, -\frac{1}{2}, -\frac{3}{2} \right}} \]