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x²(x-3) + 4(x-3) = 0
(x²+4)(x-3) = 0
Δ < 0 dla (x² + 4) więc nie ma rozwiązań
x-3 = 0
x = 3
x²(x-3) + 4(x-3) = 0
(x²+4)(x-3) = 0
Δ < 0 dla (x² + 4) więc nie ma rozwiązań
x-3 = 0
x = 3