(1/2)^(2*x)-(1/2)^x-12<0 inequation

Given the inequality:
$$\left(- \left(\frac{1}{2}\right)^{x} + \left(\frac{1}{2}\right)^{2 x}\right) - 12 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- \left(\frac{1}{2}\right)^{x} + \left(\frac{1}{2}\right)^{2 x}\right) - 12 = 0$$
Solve:
Given the equation:
$$\left(- \left(\frac{1}{2}\right)^{x} + \left(\frac{1}{2}\right)^{2 x}\right) - 12 = 0$$
or
$$\left(- \left(\frac{1}{2}\right)^{x} + \left(\frac{1}{2}\right)^{2 x}\right) - 12 = 0$$
Do replacement
$$v = \left(\frac{1}{4}\right)^{x}$$
we get
$$-12 - 2^{- x} + 2^{- 2 x} = 0$$
or
$$-12 - 2^{- x} + 2^{- 2 x} = 0$$
do backward replacement
$$\left(\frac{1}{4}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
$$x_{1} = -2$$
$$x_{2} = \frac{- \log{\left(3 \right)} + i \pi}{\log{\left(2 \right)}}$$
Exclude the complex solutions:
$$x_{1} = -2$$
This roots
$$x_{1} = -2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\left(- \left(\frac{1}{2}\right)^{x} + \left(\frac{1}{2}\right)^{2 x}\right) - 12 < 0$$

 -2*(-21)    21         
 --------    --         
    10       10         
2         - 2   - 12 < 0

        10___      5 ___    
-12 - 4*\/ 2  + 16*\/ 2  < 0
    

but

        10___      5 ___    
-12 - 4*\/ 2  + 16*\/ 2  > 0
    

Then
$$x < -2$$
no execute
the solution of our inequality is:
$$x > -2$$

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