Solve the following equation for \( x \) : \( \left(4^{2}-2 \cdot 1\right) x+3=2 x+3+x \)

To solve the equation \( \left(4^{2}-2 \cdot 1\right) x+3=2 x+3+x \), we first simplify each side. Start with the left side: \[ 4^2 = 16 \quad \text{and} \quad 2 \cdot 1 = 2 \] So, \[ 4^2 - 2 \cdot 1 = 16 - 2 = 14 \] Thus, the left side becomes: \[ 14x + 3 \] Now simplify the right side: \[ 2x + 3 + x = 3x + 3 \] Now we can rewrite the equation: \[ 14x + 3 = 3x + 3 \] Next, we'll get all the \( x \)-terms on one side and the constants on the other: \[ 14x - 3x + 3 - 3 = 0 \] This simplifies to: \[ 11x = 0 \] Now, divide both sides by 11: \[ x = 0 \] Thus, the solution to the equation is: \[ \boxed{0} \]