Two Steps With the Curvature

A one-parameter model has the loss

$L(w) = \frac{\bigl(w^{2} - 12\bigr)^{2}}{4}$

Rather than gradient descent, the optimizer uses Newton's method, which throws away the hand-chosen learning rate and reads the step size off the curvature instead:

$w_{n+1} = w_n - \frac{L'(w_n)}{L''(w_n)}$

Training starts at

$w_0 = 6$

Run exactly two Newton steps. What is $w_2$?

Round your answer to two decimal places.