How is this an extraneous solution? Let's say we've got $\sqrt{4-3x} = x$. Now let's say we have $x = 1,-4$. Let's punch $-4$ in first. We wind up with $\pm4 = -4$. > How is this an extraneous solution, but plugging in 1 isn't? Isn't a +- equal to negatives and positives?

Ok, here's the thing. A square root, when placed by itself, isn't negative, unless you put a "minus" in front of it. However, if your original equation was $x^2 = 4-3x$, both solutions would work. The confusion here arrives from you believing that $\sqrt{16}$ can be both positive and negative. Unless otherwise stated, $\sqrt{16}$ is always $4$, not $-4$, and so $4$ doesn't equal $-4$.