3+-sqrt(3x+1)=x equation

Given the equation
$$3 - \sqrt{3 x + 1} = x$$
$$- \sqrt{3 x + 1} = x - 3$$
We raise the equation sides to 2-th degree
$$3 x + 1 = \left(x - 3\right)^{2}$$
$$3 x + 1 = x^{2} - 6 x + 9$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 9 x - 8 = 0$$
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 - it is the discriminant.
Because
$$a = -1$$
$$b = 9$$
$$c = -8$$
, then

D = b^2 - 4 * a * c = 

(9)^2 - 4 * (-1) * (-8) = 49

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = 1$$
$$x_{2} = 8$$

Because
$$\sqrt{3 x + 1} = 3 - x$$
and
$$\sqrt{3 x + 1} \geq 0$$
then
$$3 - x \geq 0$$
or
$$x \leq 3$$
$$-\infty < x$$
The final answer:
$$x_{1} = 1$$