x³-3x+1 = 0

Try substituting x=0 to see if we get close to 0

0³-3·0+1 = 1
       
x=0 gives 1, too big

Try 1

1³-3·2+1 
       -1
x=1 gives -1, too small

Try something in between 0 and 1

Try .5

.5³-3·(.5)+1 = -.375 
       
x=-.375 is closer to 0, but still too small

So we try something in between 0 and .5, closer to .5, 

Try .3

.3³-3·(.3)+1 = .127 
       
.127 is closer to 0, but it's too big

So we try something in between .3 and .5, closer to .3

Try .35

.35³-3·(.35)+1 = -.007125 

Nope, that's closer to 0, but it's too small

So we try something in between .3 and .35, closer to .35

Try .345

.345³-3·(.345)+1 = .006063635        

Nope, that's closer to 0, but it's too big

So we try something in between .345 and .35,
closer to .345, 

Try .347

.347³-3·(.347)+1 = .00078192 
       
Nope, closer to 0, but still too big

So we try something in between .347 and .35,
closer to .347

Try .348

.348³-3·(.348)+1 = .0018558 
       
Nope, not closer to 0, still too big

So we try something in between .347 and .38,
closer to .347

Try .3472

.3472³-3·(.3472)+1 = .00025421 
       
Nope, closer to 0, still too big

So we try something in between .3472 and .345,
but much closer to .3472 since .00025421 is much
closer to 0 than -.007125.

So try .3473 

.3473³-3·(.3473)+1 = -.000009615

That's a lot closer to 0 than when we tried .3472

So we choose something much closer to .3473 than to .3472.

Se we can stop there.

To four decimal places, it's .3473

Edwin