sqrt(3x+28)=x equation

Given the equation
$$\sqrt{3 x + 28} = x$$
$$\sqrt{3 x + 28} = x$$
We raise the equation sides to 2-th degree
$$3 x + 28 = x^{2}$$
$$3 x + 28 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 3 x + 28 = 0$$
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
$$a = -1$$
$$b = 3$$
$$c = 28$$
, then

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

(3)^2 - 4 * (-1) * (28) = 121

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

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

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

or
$$x_{1} = -4$$
$$x_{2} = 7$$

Because
$$\sqrt{3 x + 28} = x$$
and
$$\sqrt{3 x + 28} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 7$$