Use the quadratic formula to solve. Express your answer in simplest form. \[ 4 k^{2}+6 k+1=2 k \]

To solve the quadratic equation using the quadratic formula, follow these steps: Given the equation: \[ 4k^{2} + 6k + 1 = 2k \] 1. **Rearrange the equation to standard form:** \[ 4k^{2} + 6k + 1 - 2k = 0 \\ 4k^{2} + 4k + 1 = 0 \] 2. **Identify the coefficients:** \[ a = 4, \quad b = 4, \quad c = 1 \] 3. **Apply the quadratic formula:** \[ k = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Plugging in the values: \[ k = \frac{-4 \pm \sqrt{4^{2} - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \\ k = \frac{-4 \pm \sqrt{16 - 16}}{8} \\ k = \frac{-4 \pm \sqrt{0}}{8} \\ k = \frac{-4}{8} \\ k = -\frac{1}{2} \] Since the discriminant (\(b^{2} - 4ac\)) is zero, there is exactly one real solution (a repeated root). **Final Answer:** \[ k = -\frac{1}{2} \]