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Aus Online Mathematik Brückenkurs 2

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

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

z−22−22+5=0z−22+1=0

Taking the root then gives that the equation has roots z−2=i i.e. z=2+i and 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.

z=2+i:z2−4z+5=2+i2−42+i+5=22+4i+i2−8−4i+5=4+4i−1−8−4i+5=0

z=2-i:z2−4z+5=2-i2−42-i+5=22−4i+i2−8+4i+5=4−4i−1−8+4i+5=0