$$3\sqrt{7x-5}-4=8$$
On my homework, it said, "Solve each of the following radical equations algebraically. State any restrictions on the variable." I already solved the equation algebraically and got an answer of $x = 3$. I'm not exactly sure how to state any restrictions on the variable. Any pointers in the right direction would be greatly appreciated.
Whenever you see a square root (or any even $n$-th root), the expression under that root cannot be negative (because even powers are always positive!). This means that writing down the equation $$3\sqrt{\color{blue}{7x-5}}-4=8$$ is only meaningful if $\color{blue}{7x-5} \ge 0$. This restricts the possible values of the variable $x$: $$7x-5 \ge 0 \iff 7x \ge 5 \iff x \ge \frac{5}{7}$$
Note that the solution you found, $x=3$, satisfies this condition so it is a valid solution. But you could end up with a "solution" that does not satisfy this condition and then you would discard this answer on the basis of the restriction on $x$.
You can't take the square root of a negative number, so we must have
$$7x-5\ge0$$
The $\sqrt{7x - 5}$ is only defined whenever the radicand (i.e.: $7x-5$) is non-negative. In other words,
$$ 7x - 5 \geq 0 $$
This is because, when dealing with the field of real numbers, the square root of a negative real number is not defined. As a result, the square root of a negative is actually an imaginary number involving the unit $i$.
The equation is clearly equivalent to $\sqrt{7x-5}=4$.
There is a restriction on the variable, namely $7x-5\ge0$. However, there's no need to check that after squaring the solution satisfies it, because the equation becomes $$ 7x-5=16 $$ and thus, obviously, $7x-5\ge0$.
In other cases, the condition needs to be checked. Suppose the equation is $$ \sqrt{x+3}=x+1 $$ Then the variable should satisfy $x+3\ge0$ and $x+1\ge0$, so $x\ge-1$. After squaring we get $x+3=x^2+2x+1$, hence $x^2+x-2=0$. This has the roots $1$ and $-2$; the former is good, the latter isn't.
Your answer is one solution and it is correct. Another answer, taking the value of the radical first, would be $\frac{-11}{7}$ but that conflicts with a restriction shown at the end. Without the restriction we could end up showing $4i=4$.
$$3\sqrt{7x-5}-4=8$$ $$\sqrt{7x-5}=\frac{8+4}{3}=4$$ Squaring both sides we get $$7x-5=16$$ $$x=\frac{21}{7}=3$$
If our answer is to be real we must have $7x-5\ge 0\iff7x\ge 5\iff x\ge \frac{5}{7}$ and this coincides with our solution.