Solve for $x$: $\sqrt{ x + 4} = −2$ Solve for $x$: $\sqrt{ x + 4} = −2$
I did a lot of positive numbers.
I did negative exponents.
I did everything. 
It doesn't seem possible or is it? It might be a trick question...
 A: (Answer to original question, which was mis-formatted):
You get $\sqrt{x} = -2-4 = -6.$
Since there are no real numbers whose square root is negative (the $\sqrt{\cdot}$ notation means "the positive square root or zero"), the solution set is the empty set (by which I mean that there is no solution to the problem -- there's no real number $x$ that makes the equation true). 
It's possible that your book was hoping that you'd square both sides to get
$x = 36$ and conclude that $x = 36$ is an answer, which is wrong (i.e., they were trying to fool you into making a mistake.)

OP says that the actual equation to be solved is 
$$
\sqrt{x + 4} = -2.
$$
Since $\sqrt{\cdot}$ always (by definition) produces a non-negative number, there's no solution. But your book may have hoped that you'd write
\begin{align}
\sqrt{x + 4} &= -2\\
(x + 4) &= (-2)^2\text{, by squaring both sides}\\
x + 4 &= 4\text{, because $-2 \cdot -2 = 4$}\\
x  &= 0\text{, by subtracting $4$ from both sides}\\
\end{align}
and conclude that $x = 0$ is a solution, which is not true: if you plug $x = 0$ into the left half, you get $\sqrt{0+4} = \sqrt{4} = 2$, and this is not the same as the right hand side. 
The failure was in the first line of the deduction: the solutions to an equation $A = B$ are not the same as the solutions to $A^2 = B^2$; put less formally, "it's not OK to square both sides of an equation when you're trying to solve it, or it you do, you have to check at the end that the solutions you got really do work in the original equation." 
