(3x-1)(x+3)=x(1+6x)

To solve the equation \((3x - 1)(x + 3) = x(1 + 6x)\), we will first expand both sides.

1. Expand the left side: \[ (3x - 1)(x + 3) = 3x \cdot x + 3 \cdot 3x - 1 \cdot x - 1 \cdot 3 \] \[ = 3x^2 + 9x - x - 3 \] \[ = 3x^2 + 8x - 3 \]

2. Expand the right side: \[ x(1 + 6x) = x \cdot 1 + x \cdot 6x = x + 6x^2 \]

3. Now we rewrite the equation: \[ 3x^2 + 8x - 3 = x + 6x^2 \]

4. Move all terms to one side of the equation: \[ 3x^2 + 8x - 3 - x - 6x^2 = 0 \] \[ 3x^2 - 6x^2 + 8x - x - 3 = 0 \] \[ -3x^2 + 7x - 3 = 0 \]

5. Multiply through by -1 to simplify: \[ 3x^2 - 7x + 3 = 0 \]

6. Now, we can use the quadratic formula to find solutions for \(x\): The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 3\), \(b = -7\), and \(c = 3\).

   First, calculate the discriminant: \[ b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot 3 = 49 - 36 = 13 \]

   Now apply the quadratic formula: \[ x = \frac{-(-7) \pm \sqrt{13}}{2 \cdot 3} = \frac{7 \pm \sqrt{13}}{6} \]

So, the solutions to the equation \((3x - 1)(x + 3) = x(1 + 6x)\) are: \[ x = \frac{7 + \sqrt{13}}{6} \quad \text{and} \quad x = \frac{7 - \sqrt{13}}{6} \]