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

Solve the equation \( k^{2}-4k+1=-5 \). Solve the equation(The complex numbers system) by following steps: - step0: Solve using the quadratic formula in the complex numbers system: \(k^{2}-4k+1=-5\) - step1: Move the expression to the left side: \(k^{2}-4k+6=0\) - step2: Solve using the quadratic formula: \(k=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 6}}{2}\) - step3: Simplify the expression: \(k=\frac{4\pm \sqrt{-8}}{2}\) - step4: Simplify the expression: \(k=\frac{4\pm 2\sqrt{2}\times i}{2}\) - step5: Separate into possible cases: \(\begin{align}&k=\frac{4+2\sqrt{2}\times i}{2}\\&k=\frac{4-2\sqrt{2}\times i}{2}\end{align}\) - step6: Simplify the expression: \(\begin{align}&k=2+\sqrt{2}\times i\\&k=\frac{4-2\sqrt{2}\times i}{2}\end{align}\) - step7: Simplify the expression: \(\begin{align}&k=2+\sqrt{2}\times i\\&k=2-\sqrt{2}\times i\end{align}\) - step8: Rewrite: \(k_{1}=2-\sqrt{2}\times i,k_{2}=2+\sqrt{2}\times i\) - step9: Remove the complex number(s): \(\textrm{No real solution}\) The equation \(k^{2}-4k+1=-5\) does not have any real solutions.