absolute(x^2-9x+7)=7 equation

For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x^{2} - 9 x + 7 \geq 0$$
or
$$\left(x \leq \frac{9}{2} - \frac{\sqrt{53}}{2} \wedge -\infty < x\right) \vee \left(\frac{\sqrt{53}}{2} + \frac{9}{2} \leq x \wedge x < \infty\right)$$
we get the equation
$$\left(x^{2} - 9 x + 7\right) - 7 = 0$$
after simplifying we get
$$x^{2} - 9 x = 0$$
the solution in this interval:
$$x_{1} = 0$$
$$x_{2} = 9$$

2.
$$x^{2} - 9 x + 7 < 0$$
or
$$x < \frac{\sqrt{53}}{2} + \frac{9}{2} \wedge \frac{9}{2} - \frac{\sqrt{53}}{2} < x$$
we get the equation
$$\left(- x^{2} + 9 x - 7\right) - 7 = 0$$
after simplifying we get
$$- x^{2} + 9 x - 14 = 0$$
the solution in this interval:
$$x_{3} = 2$$
$$x_{4} = 7$$


The final answer:
$$x_{1} = 0$$
$$x_{2} = 9$$
$$x_{3} = 2$$
$$x_{4} = 7$$