Question

x^2-4x+5=0的复数解x^2-4x+5=0的根为x1=2+i,x2=2-i,怎么算的?

Answer 1

x=[-b±√(b^2-4ac)]/2a

=[4±√(16-20)]/2

=[4±2√(-1)]/2

=2±√(-1)

=2±i