(x-4)^2=<4-x inequation

Given the inequality:
$$\left(x - 4\right)^{2} \leq - x + 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 4\right)^{2} = - x + 4$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(x - 4\right)^{2} = - x + 4$$
to
$$\left(x - 4\right)^{2} + \left(x - 4\right) = 0$$
Expand the expression in the equation
$$\left(x - 4\right)^{2} + \left(x - 4\right) = 0$$
We get the quadratic equation
$$x^{2} - 7 x + 12 = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = 12$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 12 + \left(-7\right)^{2} = 1$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 4$$
Simplify
$$x_{2} = 3$$
Simplify
$$x_{1} = 4$$
$$x_{2} = 3$$
$$x_{1} = 4$$
$$x_{2} = 3$$
This roots
$$x_{2} = 3$$
$$x_{1} = 4$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 3$$
=
$$\frac{29}{10}$$
substitute to the expression
$$\left(x - 4\right)^{2} \leq - x + 4$$
$$\left(\left(-1\right) 4 + \frac{29}{10}\right)^{2} \leq \left(-1\right) \frac{29}{10} + 4$$

121    11
--- <= --
100    10

but

121    11
--- >= --
100    10

Then
$$x \leq 3$$
no execute
one of the solutions of our inequality is:
$$x \geq 3 \wedge x \leq 4$$

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        /     \  
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       x_2      x_1