sqrt(1-3*x)-sqrt(x+2)=1 equation

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

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

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

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

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

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

or
$$x_{1} = -1$$
$$x_{2} = \frac{1}{4}$$

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

0 = 0

- the identity
$$x_{2} = \frac{1}{4}$$
$$\sqrt{1 - 3 x_{2}} - \sqrt{x_{2} + 2} - 1 = 0$$
=
$$\left(- \sqrt{\frac{1}{4} + 2} + \sqrt{1 - \frac{3}{4}}\right) - 1 = 0$$
=

-2 = 0

- No
The final answer:
$$x_{1} = -1$$