x-3sqrt(x)+2=0 equation

Given the equation
$$\left(- 3 \sqrt{x} + x\right) + 2 = 0$$
$$- 3 \sqrt{x} = - x - 2$$
We raise the equation sides to 2-th degree
$$9 x = \left(- x - 2\right)^{2}$$
$$9 x = x^{2} + 4 x + 4$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 5 x - 4 = 0$$
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
$$a = -1$$
$$b = 5$$
$$c = -4$$
, then

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

(5)^2 - 4 * (-1) * (-4) = 9

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

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

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

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

Because
$$\sqrt{x} = \frac{x}{3} + \frac{2}{3}$$
and
$$\sqrt{x} \geq 0$$
then
$$\frac{x}{3} + \frac{2}{3} \geq 0$$
or
$$-2 \leq x$$
$$x < \infty$$
The final answer:
$$x_{1} = 1$$
$$x_{2} = 4$$