Algebra Difference in Roots Question. 
Let D be the absolute value of the difference of the 2 roots of the equation 3x^2-10x-201=0. Find [D]. [x] denotes the greatest integer less than or equal to x.

I came across this question in a Math Competition and I am not sure how to solve it without using a calculator, since calculators are not allowed in the competition. Thanks.

$$3x^2-10x-201=0\\ \iff x^2-\frac{10}3x-67=0$$
Assuming the quadratic formula is available to use,
$$x=\frac{10}6\pm\frac{\sqrt{\frac{100}9+4\cdot 67}}2\\=\frac 53\pm\sqrt{\frac{25}9+67}$$
So the square root term is greater than $\sqrt{64}$ but less than $\sqrt{81}$ and is therefore between $8$ and $9$ in value, and therefore the difference between the roots can be either be $16$ or $17$, but not $18$ since that term is less than $9$.
The difference can be determined this way because both roots have the offset fraction $\dfrac53$ which is removed upon subtraction.
To determine whether the difference is greater than $17$, consider whether the square root term is greater than $8.5:$
$$(8+0.5)^2=64+2\cdot8\cdot0.5+0.25=67+5+0.25$$
And our original square root term contains
$$67+\frac{25}9=67+2+\frac79$$
Therefore, half of the difference between the roots is less than $8.5$ but greater than $8$ and therefore the total difference is between $16$ and $17$, leaving $16$ as the value of $[D]$.
