sqrt(2x+7)+sqrt(x+4)=sqrt(6x+18) equation

Given the equation
$$\sqrt{x + 4} + \sqrt{2 x + 7} = \sqrt{6 x + 18}$$
We raise the equation sides to 2-th degree
$$\left(\sqrt{x + 4} + \sqrt{2 x + 7}\right)^{2} = 6 x + 18$$
or
$$1^{2} \left(2 x + 7\right) + \left(2 \sqrt{\left(x + 4\right) \left(2 x + 7\right)} + 1^{2} \left(x + 4\right)\right) = 6 x + 18$$
or
$$3 x + 2 \sqrt{2 x^{2} + 15 x + 28} + 11 = 6 x + 18$$
transform:
$$2 \sqrt{2 x^{2} + 15 x + 28} = 3 x + 7$$
We raise the equation sides to 2-th degree
$$8 x^{2} + 60 x + 112 = \left(3 x + 7\right)^{2}$$
$$8 x^{2} + 60 x + 112 = 9 x^{2} + 42 x + 49$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 18 x + 63 = 0$$
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
$$a = -1$$
$$b = 18$$
$$c = 63$$
, then

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

(18)^2 - 4 * (-1) * (63) = 576

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

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

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

or
$$x_{1} = -3$$
$$x_{2} = 21$$

Because
$$\sqrt{2 x^{2} + 15 x + 28} = \frac{3 x}{2} + \frac{7}{2}$$
and
$$\sqrt{2 x^{2} + 15 x + 28} \geq 0$$
then
$$\frac{3 x}{2} + \frac{7}{2} \geq 0$$
or
$$- \frac{7}{3} \leq x$$
$$x < \infty$$
$$x_{2} = 21$$
check:
$$x_{1} = 21$$
$$\sqrt{x_{1} + 4} + \sqrt{2 x_{1} + 7} - \sqrt{6 x_{1} + 18} = 0$$
=
$$- \sqrt{18 + 6 \cdot 21} + \left(\sqrt{4 + 21} + \sqrt{7 + 2 \cdot 21}\right) = 0$$
=

0 = 0

- the identity
The final answer:
$$x_{1} = 21$$