Rozwiąż nierowność 3x^2>8x+3
3x²-8x-3>0
Δ=64+4·3·3=64+36=100
√Δ=10x₁=(8-10):6=-1/3x₂=(8+10):6=3
x∈ ( -1/3;3)
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3x²-8x-3>0
Δ=64+4·3·3=64+36=100
√Δ=10
x₁=(8-10):6=-1/3
x₂=(8+10):6=3
x∈ ( -1/3;3)