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m²+2m-3 > 0
Δ=2²+4*3=4+12=16
m.z.: m=(-2-4)/2=-3 v m=(-2+4)/2=1
Wyznaczam znaki trojmianu kwadratowego:
____-3_____1______>m
+ + 0 _ _ 0 + +
Odp. m∈(-∞, -3) u (1,+∞).