√(2x+1)-√(x-3)=√x equation

Given the equation
$$- \sqrt{x - 3} + \sqrt{2 x + 1} = \sqrt{x}$$
We raise the equation sides to 2-th degree
$$\left(- \sqrt{x - 3} + \sqrt{2 x + 1}\right)^{2} = x$$
or
$$1^{2} \cdot \left(2 x + 1\right) + \left(\left(-1\right) 2 \cdot 1 \sqrt{\left(1 x - 3\right) \left(2 x + 1\right)} + \left(-1\right)^{2} \cdot \left(1 x - 3\right)\right) = x$$
or
$$3 x - 2 \sqrt{2 x^{2} - 5 x - 3} - 2 = x$$
transform:
$$- 2 \sqrt{2 x^{2} - 5 x - 3} = 2 - 2 x$$
We raise the equation sides to 2-th degree
$$8 x^{2} - 20 x - 12 = \left(2 - 2 x\right)^{2}$$
$$8 x^{2} - 20 x - 12 = 4 x^{2} - 8 x + 4$$
Transfer the right side of the equation left part with negative sign
$$4 x^{2} - 12 x - 16 = 0$$
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
$$a = 4$$
$$b = -12$$
$$c = -16$$
, then

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

(-12)^2 - 4 * (4) * (-16) = 400

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

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

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

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

Because
$$\sqrt{2 x^{2} - 5 x - 3} = x - 1$$
and
$$\sqrt{2 x^{2} - 5 x - 3} \geq 0$$
then
$$x - 1 \geq 0$$
or
$$1 \leq x$$
$$x < \infty$$
$$x_{1} = 4$$
check:
$$x_{1} = 4$$
$$- \sqrt{x_{1}} - \sqrt{x_{1} - 3} + \sqrt{2 x_{1} + 1} = 0$$
=
$$- \sqrt{4} + \left(- \sqrt{4 - 3} + \sqrt{1 + 2 \cdot 4}\right) = 0$$
=

0 = 0

- the identity
The final answer:
$$x_{1} = 4$$