(sqrt(3*x)+4)=x equation

Given the equation
$$\sqrt{3 x} + 4 = x$$
$$\sqrt{3} \sqrt{x} = x - 4$$
We raise the equation sides to 2-th degree
$$3 x = \left(x - 4\right)^{2}$$
$$3 x = x^{2} - 8 x + 16$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 11 x - 16 = 0$$
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
$$a = -1$$
$$b = 11$$
$$c = -16$$
, then

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

(11)^2 - 4 * (-1) * (-16) = 57

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

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

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

or
$$x_{1} = \frac{11}{2} - \frac{\sqrt{57}}{2}$$
$$x_{2} = \frac{\sqrt{57}}{2} + \frac{11}{2}$$

Because
$$\sqrt{x} = \frac{\sqrt{3} x}{3} - \frac{4 \sqrt{3}}{3}$$
and
$$\sqrt{x} \geq 0$$
then
$$\frac{\sqrt{3} x}{3} - \frac{4 \sqrt{3}}{3} \geq 0$$
or
$$4 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = \frac{\sqrt{57}}{2} + \frac{11}{2}$$