11) \( k^{2}-4 k+1=-5 \)

To solve the equation \( k^{2}-4k+1=-5 \), first, we'll bring all terms to one side of the equation. This gives us: \[ k^{2} - 4k + 1 + 5 = 0 \] Simplifying this results in: \[ k^{2} - 4k + 6 = 0 \] Next, we can apply the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = 6 \). Calculating the discriminant: \[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 6 = 16 - 24 = -8 \] Because the discriminant is negative, this means there are no real solutions; instead, we will have two complex solutions. Continuing with the quadratic formula: \[ k = \frac{4 \pm \sqrt{-8}}{2 \cdot 1} = \frac{4 \pm 2i\sqrt{2}}{2} = 2 \pm i\sqrt{2} \] Thus, the solutions are: \[ k = 2 + i\sqrt{2} \quad \text{and} \quad k = 2 - i\sqrt{2} \] These solutions reveal that the equation's roots exist in the realm of complex numbers due to the negative discriminant.