(3x-1)*(3x+1)-(x-1)*(x+2)=8 x1=? x2=?

Let's solve the equation step by step:

The equation is:

\[ (3x - 1)(3x + 1) - (x - 1)(x + 2) = 8 \]

First, simplify the left side of the equation:

1. Expand \((3x - 1)(3x + 1)\): \[ (3x - 1)(3x + 1) = 9x^2 - 1 \]

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

Now substitute these expansions back into the equation:

\[ 9x^2 - 1 - (x^2 + x - 2) = 8 \]

Distribute the negative sign:

\[ 9x^2 - 1 - x^2 - x + 2 = 8 \]

Combine like terms:

\[ 9x^2 - x^2 - x - 1 + 2 = 8 \] \[ 8x^2 - x + 1 = 8 \]

Now, bring everything to one side of the equation:

\[ 8x^2 - x + 1 - 8 = 0 \] \[ 8x^2 - x - 7 = 0 \]

Now we can use the quadratic formula to solve for \(x\):

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 8\), \(b = -1\), and \(c = -7\).

First, calculate the discriminant: \[ b^2 - 4ac = (-1)^2 - 4(8)(-7) = 1 + 224 = 225 \]

Now substitute this back into the formula:

\[ x = \frac{-(-1) \pm \sqrt{225}}{2(8)} \] \[ x = \frac{1 \pm 15}{16} \]

This gives us two potential solutions:

1. \[ x_1 = \frac{1 + 15}{16} = \frac{16}{16} = 1 \]

2. \[ x_2 = \frac{1 - 15}{16} = \frac{-14}{16} = -\frac{7}{8} \]

Thus, the solutions are:

\[ x_1 = 1, \quad x_2 = -\frac{7}{8} \]