2x(3x-2)-3(x2-4x)=3x(x-7)+2

To solve the equation \(2x(3x-2)-3(x^2-4x)=3x(x-7)+2\), we will simplify both sides step-by-step.

1. Expand both sides:

   Left-hand side: \[ 2x(3x - 2) - 3(x^2 - 4x) \] First, expand \(2x(3x - 2)\): \[ 2x \cdot 3x - 2 \cdot 2x = 6x^2 - 4x \] Now, expand \(-3(x^2 - 4x)\): \[ -3x^2 + 12x \] Combining those results for the left-hand side: \[ 6x^2 - 4x - 3x^2 + 12x = (6x^2 - 3x^2) + (-4x + 12x) = 3x^2 + 8x \]

   Right-hand side: \[ 3x(x - 7) + 2 \] Expand \(3x(x - 7)\): \[ 3x^2 - 21x \] So the right-hand side becomes: \[ 3x^2 - 21x + 2 \]

2. Set both sides equal to each other: \[ 3x^2 + 8x = 3x^2 - 21x + 2 \]

3. Eliminate \(3x^2\) from both sides: \[ 8x = -21x + 2 \]

4. Add \(21x\) to both sides: \[ 8x + 21x = 2 \] \[ 29x = 2 \]

5. Solve for \(x\): \[ x = \frac{2}{29} \]

Thus, the solution to the equation is: \[ \boxed{\frac{2}{29}} \]