The way I would do it is to use a calculator. But first try substituting values of x. For example, x=0 gives y=3, which is higher than zero. Now we substitute x=1 to see if the function is getting closer to zero: y=4. Clearly it is moving away from zero, so try x=-1: y=2, which is closer to zero; so perhaps we are heading towards a zero for the function.
Now try x=-2: y=3*128-4*8+3=384-32+3=355, way too big! When x=2, y=-384+32+3=-349, way too negative!
But we have learned something: when x=1, y=4 and x=2, y=-349. So, somewhere between x=1 and x=2, there MUST be a zero, because y has gone from positive to negative.
Now we can use the calculator to find out which value of x gives y=0. So we go for the midpoint of 1 and 2 which is 1.5. When x=1.5, y=-34.76 approx. From this we can see that y is not so negative, therefore the zero is between 1 and 1.5.
So we pick x=1.25, and y=-3.49, even better. So the zero is between 1 and 1.25. We continue this way until we get y=0, or as close as possible.
You don't have to keep going for the midpoint each time. You could use numbers like 1.1, 1.2, etc., and then do as I did, drop to the next decimal place each time. Such as, 1.15, 1.16, 1.17, 1.18, 1.19, 1.20. Then 1.180, 1.181, and so on.
Continuing in this way I got 1.18039 as a fair approximation to the only real root.