Why can't you solve $0 = -5\sqrt{x+2}-2$ So i did this,
\begin{aligned}
2&=-5\sqrt{x+2},\\
\tfrac{2}{-5}&=\sqrt{x+2},\\
\tfrac{4}{25}&=x+2,\\
x&=-\tfrac{46}{25}.
\end{aligned}
Is there something wrong with my algebra?
 A: Remember that, $\sqrt{x+2}≥0$ and multiplying each side by $(-1)$  then you get,
$$-5\sqrt{x+2}-2=0 \Longleftrightarrow 5\sqrt{x+2}+2=0$$ which is impossible.
The sum of  non-negative and positive  numbers can not equal to zero.
EDIT:
Your argument is not generally correct. Because, $$a=b \not \Longleftrightarrow a^2=b^2.$$ For example,
$$-1=1 \\  (-1)^2=1^2 \\  1=1 $$ which is invalid.
A: $\sqrt{M} \ge 0$ always.  So you can't have the statement true.  If you assume the statement is true (which it can't be) then your math does nothing wrong but the result won't mean anything.
It's bit like saying solve $\frac x{0} = 7$.  That can't be true as nothing can be divided by $0$, but if you assume it is true then doing valid math $\require{cancel}$
$\frac x0 = 7$
$0\cdot \frac x0 = \cancel{0}\frac x{\cancel0} = 0\cdot 7$
$x = 0$ is a meaninless result.
This error gets hidden.
When every you have $a = M$ and you square both sides to get $a^2 = M^2$ you add in the possibility that $a = -M$ which was not an initial case.
In this case $-\frac 25 = \sqrt{x+2}$ is impossible as $\sqrt{x+2} \ge 0$.
But when you square both sides you add in the possibility $\sqrt{x+2} = \frac 25$ and because that is possible that is what you end up solving.
Once you get $x = -\frac {46}{25}$ you can check and see that it doesn't sovlve
$-5\sqrt{x+2} -2=$
$-5\sqrt{-\frac {46}{25}+2}-2=$
$-5\sqrt{-\frac {46}{25}+\frac {50}{25}} -2=$
$-5\sqrt{\frac 4{25}} - 2=$
$-5\frac 25 -2=$
$-2-2 = -4 \ne 0$.
But it is a solution to $0 = 5\sqrt{x +2} -2$.
A: When we have an equation, let's say something simple like $x+1=0$, then a number $a$ is a solution of the equation if the following logic holds:
If $x=a$, then $x+1=0$.
There is only one number for which this statement is true, namely $-1$, so that's the only solution of the equation. Essentially, if we plug in a solution, then the equation should be a true statement ($-1+1=0$ is such a true statement). Now the issue here is that this logical statement "if this, then that" starts with the solution: "If $x=a$, then [...]". But we don't usually know the solution right off the bat, we have to find it first. We usually do this be using equivalence transformations. These are ways to manipulate equations in such a way that the original equation is a true statement if and only if the transformed equation is true. For instance, two numbers are equal if and only if those same numbers decreased by 1 are also equal. If $a=b$, then also $a-1=b-1$, and if $a-1=b-1$, then $a=b$, no matter what $a$ and $b$ are. So subtracting $1$ is an equivalence transformation, which allows us to say the following:
$x+1=0$ if and only if $x+1-1=0-1$, which just says that $x=-1$.
What's the most interesting to us is the "if" part, which says that $x+1=0$ if $x=-1$. Exactly what it means for $-1$ to be a solution of the equation. The "only if" part then tells us that there is no other number $x$ could be to yield a true statement, so $-1$ is the only solution. When solving more complicated equations, we write each transformation step below the next and suppress the "if and only if", because we're lazy. But in your mind, between any two lines in your attempt to solve an equation, there should be an "if and only if" hidden.
Now back to your original equation. You didn't use equivalence transformations. Your second step already isn't one. You transform $\frac{2}{-5}=\sqrt{x+2}$ by squaring to $\frac{4}{25}=x+2$. But squaring is not an equivalence transformation. If $a=b$, then $a^2=b^2$, true. But if $a^2=b^2$, then it doesn't follow that $a=b$. So if we insert logical statements into your algebra, we could say that if $2=-5\sqrt{x+2}$, then $x=-\frac{46}{25}$. But for $-\frac{46}{25}$ to be a solution, the reverse should hold: If $x= -\frac{46}{25}$, then $2=-5\sqrt{x+2}$.
So what you showed is that if there is a solution, then it is $-\frac{46}{25}$. But you didn't show that this number is in fact a solution (and you can't, because it isn't).
A: You've incorrectly assumed $f(x)=f(y)\implies x = y$ (line 2 to line 3) which is not true when $f$ is non injective.
