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2x+3>0
2x>-3
x>-3/2
b)
x²-5x+6>0
Δ=b²-4ac=(-5)²-4*1*6=25-24=1
√Δ=1
x1=(5-1)/2=4/2=2
x2=(5+1)/2=6/2=3
(x-2)(x-3)>0
x∈(-∞;2)∨(3;+∞)