Answer
$x = 6$
Work Step by Step
$$\sqrt {2x + 4} - \sqrt {x + 3} - 1 = 0$$
$[\sqrt {2x + 4}]^{2} = [1 + \sqrt {x + 3}]^{2}$
$[2x + 4] = [1 + 2\sqrt {x + 3} + (x + 3)]$
$2x + 4 = 1 + 2\sqrt {x + 3} + x + 3$
$2x + 4 - x - 1 - 3 = 2\sqrt {x + 3}$
$x = 2\sqrt {x + 3}$
$(x)^{2} = (2\sqrt {x + 3})^{2}$
$x^{2} = 4(x+3)$
$$x^{2} - 4x - 12 = 0$$
This new equation is easily factorized since two factors of $12$ that, when substracted, yield $-4$, are $-6\times2$:
$(x-6)(x+2) = 0$
$x = 6$ or $x = -2$
Substituting these values into the original equation, we find that only $x = 6$ is a possible solution since $x = -2$ ends as ($-2 = 0$) which is not true.
