Помогите решить: √(3 х+4)+√(х-4)=2√х

Решите задачу:

$\sqrt{3x+4}+\sqrt{x-4} = 2\sqrt{x}, \\ x \geq 4, \\ (\sqrt{3x+4}+\sqrt{x-4})^2 = (2\sqrt{x})^2, \\ 3x+4+2\sqrt{(3x+4)(x-4)}+x-4 = 4x, \\ 2\sqrt{3x^2-8x-16} = 0, \\ \sqrt{3x^2-8x-16} = 0, \\ 3x^2-8x-16 = 0, \\ D_{/4}=(-4)^2-3\cdot(-16) = 64 = 8^2, \\ x_1 = \frac{4-8}{3} = -\frac{4}{3} < 4, \\ x_2 = \frac{4+8}{3} = 4. \\ x=4.$