Ridders' method

Remarks

Claim 1: since $f(x_0) f(x_2) < 0$, the radicand is larger than $f(x_1)^2$, so the fraction has absolute value less than $1$ and $\abs{x_3 - x_1} < x_1 - x_0$. Claim 3: $f(x_3)$ has the sign opposite to $f(x_0)$ or to $f(x_2)$, unless it is $0$. So the bracket shrinks at each step and a zero stays inside it, by the intermediate value theorem. The formula comes from a trick: choose $Q$ so that the three points of $h(x) = f(x) e^{Qx}$ at $x_0, x_1, x_2$ lie on a line, then apply regula falsi to $h$. For $f(x) = x - c$ on $[0, 2]$ with $0 < c < 2$, the radicand is $1$ and the formula gives $x_3 = c$ at once. For a smooth $f$ with a simple zero, the convergence is quadratic per step, and each step costs two evaluations of $f$.

Going deeper

For a student

The method is as safe as bisection but usually much faster. It needs no derivatives, and it is a simple alternative to Brent's method when a bracket is known. Its cost is two function values per step, so its order per function evaluation is $\sqrt{2}$.