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z²+(3+i)z+4=0
PAMIĘTAJ,ŻE i²=-1
Δ=(3+i)²-4×1×4=9+6i+i²-16=6i-8
√Δ=√6i-8√(1+3i)² =1+3i
z₁=(-3-i-1-3i)/2
z₁=(-4-4i)/2
z₁=-2-2i
z₂=(-3-i+1+3i)/2
z₂=(-2+2i)/2
z₂=-1+i