x-7sqrt(x)+10=0 equation

Given the equation
$$\left(- 7 \sqrt{x} + x\right) + 10 = 0$$
$$- 7 \sqrt{x} = - x - 10$$
We raise the equation sides to 2-th degree
$$49 x = \left(- x - 10\right)^{2}$$
$$49 x = x^{2} + 20 x + 100$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 29 x - 100 = 0$$
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
$$a = -1$$
$$b = 29$$
$$c = -100$$
, then

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

(29)^2 - 4 * (-1) * (-100) = 441

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

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

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

or
$$x_{1} = 4$$
$$x_{2} = 25$$

Because
$$\sqrt{x} = \frac{x}{7} + \frac{10}{7}$$
and
$$\sqrt{x} \geq 0$$
then
$$\frac{x}{7} + \frac{10}{7} \geq 0$$
or
$$-10 \leq x$$
$$x < \infty$$
The final answer:
$$x_{1} = 4$$
$$x_{2} = 25$$