Wondering if i Did it right
1)
(x-3)/(x+4)≤0
x+4≤0
x≤-4
(-infin,-4]
2)
4x^2+2x+3=0
x=-2±√(4-48)/8
x=-2±√(-44)/8
x=-2±√(11*4)i/8
x=-2±2√(11)i/8
x=-1/4±√(11)i/4
for the first problem, x+4 cannot equal zero. So you have $\displaystyle (x-3)\le 0 \text{ or } x+4<0$
2.) $\displaystyle \sqrt{-44} = i\sqrt{44}$ - just realised you already did that...was having trouble reading the expression, and it's correct. You can however simplify further, by putting it as a single fraction over four.
Here is another attempt:
$\displaystyle \dfrac{x-3}{x+4}\leq0~,~x \neq -4$ . The LHS is a negative quotient. A quotient is negative if numerator and denominator have different signs. That means you have:
$\displaystyle x-3\leq 0\ \wedge \ x+4>0~\vee~x-3 \geq 0\ \wedge \ x+4< 0$
$\displaystyle x\leq 3\ \wedge \ x>-4~\vee~x \geq 3\ \wedge \ x< -4$
$\displaystyle \boxed{-4<x\leq3}~\vee~ x \notin \mathbb{R}$
Sorry, yeh that should be plus three, and the intersection of the two is empty as you say, but I suppose you can look at it as not being defined for real variables at all, since for any x belongng to the intersection, the function isn't defined, or you can look at it that even though some real variables don't give a valid output, there is a domain (a subset) of the real numbers (the union) for which the function is defined. Well my opinion is that both points are as valid as the other, but maybe some situations might be better stated with one definition over the other.