√(x+3)+√(3*x-2)=7 equation

Given the equation
$$\sqrt{x + 3} + \sqrt{3 x - 2} = 7$$
We raise the equation sides to 2-th degree
$$\left(\sqrt{x + 3} + \sqrt{3 x - 2}\right)^{2} = 49$$
or
$$1^{2} \left(3 x - 2\right) + \left(2 \sqrt{\left(x + 3\right) \left(3 x - 2\right)} + 1^{2} \left(x + 3\right)\right) = 49$$
or
$$4 x + 2 \sqrt{3 x^{2} + 7 x - 6} + 1 = 49$$
transform:
$$2 \sqrt{3 x^{2} + 7 x - 6} = 48 - 4 x$$
We raise the equation sides to 2-th degree
$$12 x^{2} + 28 x - 24 = \left(48 - 4 x\right)^{2}$$
$$12 x^{2} + 28 x - 24 = 16 x^{2} - 384 x + 2304$$
Transfer the right side of the equation left part with negative sign
$$- 4 x^{2} + 412 x - 2328 = 0$$
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
$$a = -4$$
$$b = 412$$
$$c = -2328$$
, then

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

(412)^2 - 4 * (-4) * (-2328) = 132496

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

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

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

or
$$x_{1} = 6$$
$$x_{2} = 97$$

Because
$$\sqrt{3 x^{2} + 7 x - 6} = 24 - 2 x$$
and
$$\sqrt{3 x^{2} + 7 x - 6} \geq 0$$
then
$$24 - 2 x \geq 0$$
or
$$x \leq 12$$
$$-\infty < x$$
$$x_{1} = 6$$
check:
$$x_{1} = 6$$
$$\sqrt{x_{1} + 3} + \sqrt{3 x_{1} - 2} - 7 = 0$$
=
$$-7 + \left(\sqrt{3 + 6} + \sqrt{-2 + 3 \cdot 6}\right) = 0$$
=

0 = 0

- the identity
The final answer:
$$x_{1} = 6$$