Finding solution of a rational expression

Equations with a variable in a denominator or under a root often give more candidate solutions than are actually valid.
The extra ones are called extraneous, and the final step is always to check each candidate in the original equation.

A candidate is invalid if it makes any denominator zero.
Solving $\frac{2x^2 - 18}{x - 3} = 12$ leads to $x = 3$, but $x = 3$ makes $x - 3 = 0$, so the original expression is undefined there.
That value is extraneous, and here it leaves the equation with no valid solution.

A candidate is also invalid if it makes a square root equal a negative number.
For $\sqrt{3x + 1} = x - 3$, squaring gives candidates $x = 1$ and $x = 8$.
At $x = 1$ the right side is $-2$, but a square root cannot be negative, so $x = 1$ is extraneous; only $x = 8$ works, since $\sqrt{25} = 5 = 8 - 3$.

So solve the equation as usual, but treat the answers as candidates, not final.
Substitute each one back into the original, and discard any that break a denominator or a root.
Keeping only the survivors gives the true solution set, which may be one value, several, or none.

Solve $\frac{x^2 - 25}{x - 5} = 10$.
Simplifying, $\frac{(x - 5)(x + 5)}{x - 5} = x + 5 = 10$, so $x = 5$, but $x = 5$ makes the denominator $x - 5 = 0$.
That solution is extraneous, so the equation has no valid solution.

Solve $\sqrt{2x + 3} = x$.
Squaring gives $2x + 3 = x^2$, so $x^2 - 2x - 3 = 0$ with candidates $x = 3$ and $x = -1$; test each.
At $x = -1$ the root would equal $-1$, which is impossible, so it is extraneous; $x = 3$ gives $\sqrt{9} = 3$, so the only solution is $x = 3$.

The equation $\sqrt{x + 11} = x - 1$ gives candidates $x = 5$ and $x = -2$. Which is valid?
At $x = -2$ the right side is $-3$, but $\sqrt{9} = 3$, so $x = -2$ is extraneous.
At $x = 5$, $\sqrt{16} = 4 = 5 - 1$, so the valid solution is $x = 5$.