Another way of approximating this zero could combine values f(2.23) ~ -0.48489759 below the x axis and f(2.24) ~ 0.31640576 above the x axis to imply that you crossed the x axis. Check the average x = (2.23 + 2.24)/2 = 2.235 in f(x). That is, f(2.235) ~ -0.085640699375 which is also below the x axis. So try the average of x = (2.235 + 2.24)/2 = 2.2375 in f(x). Notice we take an average of the x giving f(x) below and the x giving f(x) above to see better approximations for the correct number for x. this last x gives f(2.2375)~ 0.115033227539. So back we go to the average x = (2.235 + 2.2375)/2 = 2.23625 and f(2.23625) ~ .014609014263 and with that closer value we let you make the next several steps to see that with your calculator and this method you will converge to the Zero.

But would this method have worked around x = -2? There does not exist a value of f(x) above the x axis in a small interval around x = -2. Suppose you started estimating near x = -2.04 where f(-2.04) ~ -0.00134144 and f '(-2.04) ~ 0.060544. The tangent line through this point (x, f(x)) would have and equation y = f '(-2.04)(x + 2.04) + f(-2.04). This means y = 0.060544 x + 0.12216832. If we used tangent line y to predict where f(x) hits the x axis then we would be asking y to be zero. For 0.060544 x + 0.12216832 = 0, x = -0.12216832/0.060544 = -2.0178435518 which is much closer to the answer than -2.04 and we can do the same process with our new estimate of the zero. This method is called the Newton or Newton-Raphson method which is summarized by xnew = xold - [f '(xold)/f(xold)] iterated over and over.