Aus Online Mathematik Brückenkurs 2

Typically, one solves a second-degree by completing the square, followed by taking the square root.

If we complete the square of the left-hand side, we get

\displaystyle \begin{align} (z-2)^2-2^2+5&=0,\\[5pt] (z-2)^2+1&=0. \end{align}

Taking the square root then gives that the equation has roots \displaystyle z-2=\pm i, i.e. \displaystyle z=2+i and \displaystyle z=2-i.

If we want to be sure that we have found the correct solutions, we can substitute each solution into the equation and see whether the equation is satisfied.

\displaystyle \begin{align} z=2+i:\qquad z^2-4z+5 &= (2+i)^2-4(2+i)+5\\[5pt] &= 2^2+4i+i^2-8-4i+5\\[5pt] &= 4+4i-1-8-4i+5\\[5pt] &=0\,,\\[10pt] z={}\rlap{2-i:}\phantom{2+i:}{}\qquad z^2-4z+5 &= (2-i)^2-4(2-i)+5\\[5pt] &= 2^2-4i+i^2-8+4i+5\\[5pt] &= 4-4i-1-8+4i+5\\[5pt] &= 0\,\textrm{.} \end{align}