2^x^2-3x*5^x+2<=1024 inequation

Given the inequality:
$$\left(2^{x^{2}} - 5^{x} 3 x\right) + 2 \leq 1024$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2^{x^{2}} - 5^{x} 3 x\right) + 2 = 1024$$
Solve:
$$x_{1} = 3.43929121696252$$
$$x_{2} = -3.16181860949899$$
$$x_{1} = 3.43929121696252$$
$$x_{2} = -3.16181860949899$$
This roots
$$x_{2} = -3.16181860949899$$
$$x_{1} = 3.43929121696252$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-3.16181860949899 + - \frac{1}{10}$$
=
$$-3.26181860949899$$
substitute to the expression
$$\left(2^{x^{2}} - 5^{x} 3 x\right) + 2 \leq 1024$$
$$2 + \left(- \frac{\left(-3.26181860949899\right) 3}{5^{3.26181860949899}} + 2^{\left(-3.26181860949899\right)^{2}}\right) \leq 1024$$

1597.18396443785 <= 1024

but

1597.18396443785 >= 1024

Then
$$x \leq -3.16181860949899$$
no execute
one of the solutions of our inequality is:
$$x \geq -3.16181860949899 \wedge x \leq 3.43929121696252$$

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