(2x-1)(2x+1)+x(x-1)=2x(x+1)

To solve the equation \((2x-1)(2x+1) + x(x-1) = 2x(x+1)\), we will first expand and simplify both sides.

1. Expand the left side:

   \((2x-1)(2x+1)\) is a difference of squares: \[ (2x-1)(2x+1) = (2x)^2 - (1)^2 = 4x^2 - 1 \]

   Now, expand \(x(x-1)\): \[ x(x-1) = x^2 - x \]

   So, the left side becomes: \[ 4x^2 - 1 + x^2 - x = 5x^2 - x - 1 \]

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

3. Set the equation: Now we rewrite the entire equation: \[ 5x^2 - x - 1 = 2x^2 + 2x \]

4. Move all terms to one side: Subtract \(2x^2 + 2x\) from both sides: \[ 5x^2 - x - 1 - 2x^2 - 2x = 0 \] Simplifying this gives: \[ 3x^2 - 3x - 1 = 0 \]

5. Solve the quadratic equation: Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 3\), \(b = -3\), and \(c = -1\).

   First, calculate the discriminant: \[ b^2 - 4ac = (-3)^2 - 4(3)(-1) = 9 + 12 = 21 \]

   Now plug the values into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{21}}{2 \cdot 3} = \frac{3 \pm \sqrt{21}}{6} \]

6. Final Solution: Thus, the solutions for \(x\) are: \[ x = \frac{3 + \sqrt{21}}{6} \quad \text{and} \quad x = \frac{3 - \sqrt{21}}{6} \]

These are the final solutions to the equation \((2x-1)(2x+1) + x(x-1) = 2x(x+1)\).