Solve log 1/2 (2x^2 + x + 1) >= -1

Because the logarithm has a subunit base, namely 1/2, the
logarithmic function is decreasing.

We'll re-write the
equation:

 log 1/2 (2x^2 + x + 1) >=
-1*1

log 1/2 (2x^2 + x + 1) >= -1*log 1/2
(1/2)

log 1/2 (2x^2 + x + 1) >= log 1/2
(1/2)^-1

log 1/2 (2x^2 + x + 1) >=  log 1/2
(2)

2x^2 + x + 1 =<
2

We'll subtract 2 both
sides:

2x^2 + x + 1 - 2
=< 0

2x^2 + x - 1 =<
0

We'll calculate the roots of the
equation:

2x^2 + x - 1 =
0

We'll apply the quadratic
formula:

x1 = [-1 + sqrt(1 + 8)] /
4

x1 = (-1 + 3) / 4

x1 =
2/4

x1 = 1/2

x2 = (-1 - 3) /
4

x2 = -1

The
expression is negative over the interval [-1 , 1/2] and is positive for x< -1 and
x> 1/2.