To solve the equation, we can simplify it by factoring the expressions inside the square roots:

căn(x^2 - 9) + căn(x^2 - 4x + 3) = 0

The expression inside the first square root, x^2 - 9, can be factored as (x - 3)(x + 3).

The expression inside the second square root, x^2 - 4x + 3, can be factored as (x - 1)(x - 3).

Now we can rewrite the equation:

căn((x - 3)(x + 3)) + căn((x - 1)(x - 3)) = 0

Since both square roots have a common factor of (x - 3), we can simplify further:

căn(x - 3) * (căn(x + 3) + căn(x - 1)) = 0

Now we have two cases to consider:

1) căn(x - 3) = 0
This implies x - 3 = 0
Solving for x, we get x = 3.

2) căn(x + 3) + căn(x - 1) = 0
This case requires further simplification. We can square both sides of the equation to eliminate the square roots:

(căn(x + 3) + căn(x - 1))^2 = 0^2
(căn(x + 3))^2 + 2 * căn(x + 3) * căn(x - 1) + (căn(x - 1))^2 = 0
x + 3 + 2 * căn(x + 3) * căn(x - 1) + x - 1 = 0
2x + 2 * căn(x + 3) * căn(x - 1) + 2 = 0
x + căn(x + 3) * căn(x - 1) + 1 = 0

Now we can substitute u = căn(x + 3) * căn(x - 1):

x + u + 1 = 0
x = -u - 1

However, we need to check if this solution is valid. Since u = căn(x + 3) * căn(x - 1), we can substitute it back into the equation:

u = căn((-u - 1) + 3) * căn((-u - 1) - 1)
u = căn(2 - u) * căn(-u - 2)

Squaring both sides to eliminate the square roots:

u^2 = (2 - u) * (-u - 2)
u^2 = -2u - 4 + 2u + 4
u^2 = 0

Therefore, u = 0.

Substituting u = 0 back into x = -u - 1:

x = -0 - 1
x = -1

So the solutions to the equation are x = 3 and x = -1.