x-5*sqrt(x)+6=0 equation

Given the equation
$$\left(- 5 \sqrt{x} + x\right) + 6 = 0$$
$$- 5 \sqrt{x} = - x - 6$$
We raise the equation sides to 2-th degree
$$25 x = \left(- x - 6\right)^{2}$$
$$25 x = x^{2} + 12 x + 36$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 13 x - 36 = 0$$
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
$$a = -1$$
$$b = 13$$
$$c = -36$$
, then

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

(13)^2 - 4 * (-1) * (-36) = 25

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

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

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

or
$$x_{1} = 4$$
$$x_{2} = 9$$

Because
$$\sqrt{x} = \frac{x}{5} + \frac{6}{5}$$
and
$$\sqrt{x} \geq 0$$
then
$$\frac{x}{5} + \frac{6}{5} \geq 0$$
or
$$-6 \leq x$$
$$x < \infty$$
The final answer:
$$x_{1} = 4$$
$$x_{2} = 9$$