The easiest way to solve this is to substitute some trial values for x. Put x=0 and f(x) becomes 6, a positive number. Can we find a value of x to make it negative? Since there's an x^3 term, if x is negative then so is x^3. The more negative x is the bigger the chance of getting a negative value for f(x). Putting x=-1 isn't negative enough, so let's try x=-2. Nope, f(x) becomes -8+8+6=6, because -4(-2)=+8. How about -3? This time f(x) is -27+12+6=-9. So we know that f(x) goes from positive to negative between x=-2 and x=-3. Nearly there, because to go from + to - we must pass through 0. You need your calculator for the next bit. Put x=-2.1 and evaluate f(x), then x=-2.2 and evaluate. Write down your results. Eventually f(x) will go from + to -. Here's what you should get:

f(-2.1)=5.139, f(-2.2)=4.152, f(-2.3)=3.033, f(-2.4)=1.776, f(-2.5)=0.375, f(-2.6)=-1.176. Stop! To one place of decimals (one tenth) x=-2.5 is the zero of the function. f(-2.5) gives a number closer to zero than f(-2.6).