To solve the equation
\[ \frac{3x}{16x + 12} = \frac{1}{x}, \]
we can cross-multiply to eliminate the fractions. This gives us:
\[ 3x \cdot x = 1 \cdot (16x + 12). \]
Simplifying both sides yields:
\[ 3x^2 = 16x + 12. \]
Next, we can rearrange this into a standard quadratic equation form:
\[ 3x^2 - 16x - 12 = 0. \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 3 \), \( b = -16 \), and \( c = -12 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 3 \cdot (-12) = 256 + 144 = 400. \]
Now we can plug the values into the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{400}}{2 \cdot 3} = \frac{16 \pm 20}{6}. \]
This gives us two potential solutions:
- \( x = \frac{16 + 20}{6} = \frac{36}{6} = 6 \)
- \( x = \frac{16 - 20}{6} = \frac{-4}{6} = -\frac{2}{3} \)
The solutions are \( x = 6 \) and \( x = -\frac{2}{3} \).
Check the Solutions:
- Check \( x = 6 \):
\[ \frac{3(6)}{16(6) + 12} = \frac{18}{96 + 12} = \frac{18}{108} = \frac{1}{6}, \] and \[ \frac{1}{6} = \frac{1}{6} \quad \text{(True)}. \]
- Check \( x = -\frac{2}{3} \):
\[ \frac{3(-\frac{2}{3})}{16(-\frac{2}{3}) + 12} = \frac{-2}{-\frac{32}{3} + 12} = \frac{-2}{-\frac{32}{3} + \frac{36}{3}} = \frac{-2}{\frac{4}{3}} = -2 \cdot \frac{3}{4} = -\frac{3}{2}, \] and \[ \frac{1}{-\frac{2}{3}} = -\frac{3}{2} \quad \text{(True)}. \]
Thus, both solutions are valid.
The final solutions are:
\[ \boxed{6} \quad \text{and} \quad \boxed{-\frac{2}{3}}. \]