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(125+x³)(x²-x-6)=0
x³=-125
x₁=-5
x²-x-6=0
Δ=b²-4ac=1+24=25
√Δ=5
x₂=(-b-√Δ):2a=(1-5):2=-2
x₃=(-b+√Δ):2a=(1+5):2=3
2)
2x³-6x²+8x-24=0
2x²(x-3)+8(x-3)=
(2x²+8)(x-3)
2x²+8=0/:2
x²=-4
brak pierwiastków
x-3=0
x=3
rozwiazaniem jest x=3