√5x-1=√3x-2-√2x-3 equation

Given the equation
$$\sqrt{5 x} - 1 = \left(- \sqrt{2 x} + \left(\sqrt{3 x} - 2\right)\right) - 3$$
Transfer the right side of the equation left part with negative sign
$$\sqrt{x} \left(- \sqrt{3} + \sqrt{2} + \sqrt{5}\right) = -4$$
We raise the equation sides to 2-th degree
$$x \left(- \sqrt{3} + \sqrt{2} + \sqrt{5}\right)^{2} = 16$$
$$x \left(- \sqrt{3} + \sqrt{2} + \sqrt{5}\right)^{2} = 16$$
Transfer the right side of the equation left part with negative sign
$$x \left(- \sqrt{3} + \sqrt{2} + \sqrt{5}\right)^{2} - 16 = 0$$
Expand brackets in the left part

-16 + xsqrt+2 + sqrt5 - sqrt3)^2 = 0

Looking for similar summands in the left part:

-16 + x*(sqrt(2) + sqrt(5) - sqrt(3))^2 = 0

Move free summands (without x)
from left part to right part, we given:
$$x \left(- \sqrt{3} + \sqrt{2} + \sqrt{5}\right)^{2} = 16$$
Divide both parts of the equation by (sqrt(2) + sqrt(5) - sqrt(3))^2

x = 16 / ((sqrt(2) + sqrt(5) - sqrt(3))^2)

We get the answer: x = 16/(sqrt(2) + sqrt(5) - sqrt(3))^2

Because
$$\sqrt{x} = - \frac{4}{- \sqrt{3} + \sqrt{2} + \sqrt{5}}$$
and
$$\sqrt{x} \geq 0$$
then
$$- \frac{4}{- \sqrt{3} + \sqrt{2} + \sqrt{5}} \geq 0$$
The final answer:
This equation has no roots