x²-(4-i)x+5-5i=0

x²-(4-i)x+5-5i=0
x²-(4-i)x+5-5i=0

x²-(4-i)x+5-5i=0
方程：x²-(4-i)x+5-5i=0 ,
利用x={-b±√[(b)²-4ac]}/(2a)得此方程的
x1={(4-i)+√[(4-i)²-4(5-5i)]}/2=2-(i/2)+√[3i-(5/4)]
x2={(4-i)-√[(4-i)²-4(5-5i)]}/2=2-(i/2)-√[3i-(5/4)]
又有：
√[3i-(5/4)]=1+(3/2)i,分别代入到x1和x2的方程中得：
x1=2-(i/2)+1+(3/2)i=3+i.
x2=2-(i/2)-[1+(3/2)i]=1-2i.