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(x+3)(x+5)=0
x+3=0 v x+5=0
x₁=-3 v x₂=-5
II sposób:
(x+3)(x+5)=0
x²+5x+3x+15=0
x²+8x+15=0
a=1 b=8 c=15
Δ=b²-4*a*c
Δ=8²-4*1*15
Δ=64-60
Δ=4, Δ>0, więc ma 2 miejsca zerowe:
x₁=-b-√Δ/2*a
x₁=-8-√4/2*1
x₁=-8-2/2
x₁=-10/2
x₁=-5
x₂=-b+√Δ/2*a
x₂=-8+√4/2*1
x₂=-8+2/2
x₂=-6/2
x₂=-3