Answer
$x=7, 3$
Work Step by Step
$$\sqrt {2x-5} - \sqrt {x-3} = 1$$
$$(\sqrt {2x-5})^{2} = (1 + \sqrt {x-3})^{2}$$
$$(2x-5) = 1 + 2\sqrt {x-3} + (x-3)$$
$$2x-5 - 1 - x + 3 = 2\sqrt {x-3}$$
$$(x - 3)^{2} = (2\sqrt {x - 3})^{2}$$
$$x^{2} - 6x + 9 = 4(x - 3)$$
$$x^{2} - 6x + 9 - 4x + 12 = 0$$
$$x^{2} -10x + 21 = 0$$
where two factors of 21 that, when added, give 10, are:
$$(x-7)(x-3) = 0$$
$x = 7$ or $x = 3$
