sqrt3-x=sqrt3x+1 equation

Given the equation
$$- x + \sqrt{3} = \sqrt{3 x} + 1$$
Transfer the right side of the equation left part with negative sign
$$- \sqrt{3} \sqrt{x} = x - \sqrt{3} + 1$$
We raise the equation sides to 2-th degree
$$3 x = \left(x - \sqrt{3} + 1\right)^{2}$$
$$3 x = x^{2} - 2 \sqrt{3} x + 2 x - 2 \sqrt{3} + 4$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + x + 2 \sqrt{3} x - 4 + 2 \sqrt{3} = 0$$
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
$$a = -1$$
$$b = 1 + 2 \sqrt{3}$$
$$c = -4 + 2 \sqrt{3}$$
, then

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

(1 + 2*sqrt(3))^2 - 4 * (-1) * (-4 + 2*sqrt(3)) = -16 + (1 + 2*sqrt(3))^2 + 8*sqrt(3)

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

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

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

or
$$x_{1} = - \frac{\sqrt{-16 + 8 \sqrt{3} + \left(1 + 2 \sqrt{3}\right)^{2}}}{2} + \frac{1}{2} + \sqrt{3}$$
$$x_{2} = \frac{1}{2} + \sqrt{3} + \frac{\sqrt{-16 + 8 \sqrt{3} + \left(1 + 2 \sqrt{3}\right)^{2}}}{2}$$

Because
$$\sqrt{x} = - \frac{\sqrt{3} x}{3} - \frac{\sqrt{3} \left(1 - \sqrt{3}\right)}{3}$$
and
$$\sqrt{x} \geq 0$$
then
$$- \frac{\sqrt{3} x}{3} - \frac{\sqrt{3} \left(1 - \sqrt{3}\right)}{3} \geq 0$$
or
$$-\infty < x$$
$$x \leq \frac{\sqrt{3} \left(3 - \sqrt{3}\right)}{3}$$
The final answer:
$$x_{1} = - \frac{\sqrt{-16 + 8 \sqrt{3} + \left(1 + 2 \sqrt{3}\right)^{2}}}{2} + \frac{1}{2} + \sqrt{3}$$