Xét Δ=(5−i)2−4(8−i)=−8−6i⇒δ=√Δ=±(1−3i)$\begin{cases}z_1= \dfrac{5-i+\delta}{2}=3-2i\\ z_2=\dfrac{5-i-\delta}{2}=2+i \end{cases}$
Xét
Δ=(5−i)2−4(8−i)=−8−6i$\Rightarrow \delta
_{1,2}=\sqrt{\Delta}=\pm(1-3i)$$\begin{cases}z_1= \dfrac{5-i+\delta
_1}{2}=3-2i\\ z_2=\dfrac{5-i
+\delta
_2}{2}=2+i \end{cases}$