(2x-1) (x+2) < 0

since the value is negative then we have two cases:

case (1):

2x-1 <0 AND x+ 2 > 0

x < 1/2 AND x >-2

==> -2 < x < 1/2

==> x belongs to the interval (-2, 1/2)

Case 2:

2x-1 > 0 AND x+2 < 0

x > 1/2 AND x < -2

==> x = (1/2, inf) + (-inf, -2) = empty set.

Then the solution is :

**x belongs to ( -2, 1/2)**

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