(3x-1)*(3x+1)-(x-1)*(x+2) x1=? x2=?

To solve the equation \((3x - 1)(3x + 1) - (x - 1)(x + 2) = 0\), we first expand both parts.

1. Expand \((3x - 1)(3x + 1)\): \[ (3x - 1)(3x + 1) = 9x^2 - 1 \] This uses the difference of squares formula: \(a^2 - b^2 = (a-b)(a+b)\).

2. Expand \((x - 1)(x + 2)\): \[ (x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2 \]

3. Now substitute the expanded forms back into the equation: \[ 9x^2 - 1 - (x^2 + x - 2) = 0 \]

4. Distribute the negative sign: \[ 9x^2 - 1 - x^2 - x + 2 = 0 \]

5. Combine like terms: \[ 9x^2 - x^2 - x - 1 + 2 = 0 \\ 8x^2 - x + 1 = 0 \]

Now, we can solve the quadratic equation \(8x^2 - x + 1 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 8\), \(b = -1\), and \(c = 1\).

Calculating the discriminant: \[ b^2 - 4ac = (-1)^2 - 4 \cdot 8 \cdot 1 = 1 - 32 = -31 \]

Since the discriminant is negative, this indicates there are no real solutions for \(x\). However, we can still express the solutions in terms of complex numbers: \[ x = \frac{1 \pm \sqrt{-31}}{16} \\ x = \frac{1 \pm i\sqrt{31}}{16} \]

Thus, the solutions are: \[ x_1 = \frac{1 + i\sqrt{31}}{16} \\ x_2 = \frac{1 - i\sqrt{31}}{16} \]