I have a radical equation to solve:
Before checking for extraneous solutions I arrived at -4 and 3. My textbook says the only solution is 3 but surely it's -4 too?
This is correct no? A square root of 16 is -4?
$12-x=x^2$ # remove radical by squaring both sides
$-x^2-x+12=0$ # move everything to one side
$x^2+x-12=0$ # multiple both sides by -1 to get a positive exponential term
$x^2+4x-3x-12=0$ # split into groups (what's the conventional name of this step?)
$x(x+4)-3(x+4)=0$ # not sure the name of this step? "pre" factoring?
$(x+4)(x-3)$ # factor into groups
Why is it that 3 is the only solution?