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2 -3 -8 12
2 0 4 2 -12
2 1 -6 =
czyli mamy (x-2)(2x^2+x-6)=0
x-2=0 lub 2x^2+x-6=0
x=2 lub ( Δ=49 ) x=-2 lub x=1,5
x∈{-2;1,5:2}
2.
1 2 2 -5
1 0 1 3 5
1 3 5 =
(x-1)(x^2+3x+5)=0
x=1 ∨ x^2+3x+5≠0 bo Δ=-11
x∈{1}