Getting rid of the square root

A square root is undone by squaring, because $(\sqrt{a})^2 = a$. To keep the equation balanced you must square both sides, not just the side with the root.

Isolate the radical before squaring, or the root will not disappear cleanly. For $4 - \sqrt{5x - 10} = 2x$, first move the root by itself: $\sqrt{5x - 10} = 4 - 2x$. Then square both sides: $5x - 10 = (4 - 2x)^2$.

Squaring usually turns the equation into a quadratic to solve. Continuing above, $(4 - 2x)^2 = 16 - 16x + 4x^2$, so $5x - 10 = 4x^2 - 16x + 16$, which rearranges to $4x^2 - 21x + 26 = 0$.

If there are two square-root terms added together, isolate one and square to clear it, then repeat if a root remains. Squaring the whole thing while the terms are added leaves a stubborn cross term with a root still in it.

Turn $\sqrt{x + 11} = x - 1$ into a polynomial equation.
Square both sides: $(\sqrt{x + 11})^2 = (x - 1)^2$.
This gives $x + 11 = x^2 - 2x + 1$.

Express $5 - \sqrt{2x + 3} = 2x$ in the form $ax^2 + bx + c = 0$.
Isolate the root: $\sqrt{2x + 3} = 5 - 2x$.
Square: $2x + 3 = 25 - 20x + 4x^2$, so $4x^2 - 22x + 22 = 0$.

Why not square $3\sqrt{2}\,x - \sqrt{ax} = 0$ right away?
Squaring while the two terms are subtracted leaves a cross term that still has a root. Instead, move one term over first: $3\sqrt{2}\,x = \sqrt{ax}$.
Now square both sides: $18x^2 = ax$, which is free of roots.