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a+b+c = 0
b = -a-c
Δ=b²-4ac ≥ 0
(-a-c)² -4ac ≥ 0
a²+2ac + c -4ac ≥ 0
a²-2ac + c ≥ 0
(a-c)² ≥ 0
∀a,c∈R.(a-c)² ≥ 0
więc równanie to ma zawsze co najmniej 1 pierwiastek