Question

z^2+(5-2i)z+5(1-i)=0

Answer 1

$z^2+(5-2i)z+5(1-i)=0$

$D=(5-2i)^2-4\cdot5(1-i)=25-20i+4i^2-20+20i=25-4-20=1$

$z=\dfrac{2i-5\pm1}{2}$

$z_1=\dfrac{2i-5-1}{2} =\dfrac{2i-6}{2} =i-3$

$z_2=\dfrac{2i-5+1}{2} =\dfrac{2i-4}{2} =i-2$

Ответ: i-3; i-2