trial and improvement

not a nice question but it does highlight a need to work to two decimal places if asked to give an answer to 1 d.p.

between 2 and 3 find a solution to 1 d.p. to:

substituting,

x = 2.1 gives y = -3.969

x = 2.2 gives y = -3.872

x = 2.1 is just slightly closer to the required y value (of -3.921)
being 0.048 off it compared with 0.049
since it is nearer then it might be tempting to say the solution to 1 d.p. is 2.1

but, the solution to 1 d.p is actually 2.2 and not 2.1

substituting x = 2.15 (as required in an exam question) gives y = -3.929125 so the answer is bigger than 2.15 (actually around 2.158...) which rounds to 2.2

moral: don't jump to conclusions when you are dealing with non-linear functions

exam boards sensibly steer clear of cubic functions with an x squared term

for (another un-nice question) between 3 and 4 find a solution to 1 d.p. to:

then you can obtain two solutions
x = 3.3 and x = 3.4 are both correct

(actual values are 3.29188523551653 and 3.3744064395661)