If the mean of the data Class 5-10 10-15 15-20 20-25 25-30 30-35...

The class marks (mid-points) $x_i$ for the given classes are $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$.

The sum of the frequencies is:
$\sum f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 2k + 90$

The sum of the products $f_i x_i$ is:
$\sum f_i x_i = 2(7.5) + k(12.5) + 28(17.5) + 54(22.5) + (k + 1)(27.5) + 5(32.5)$

$\sum f_i x_i = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5$

$\sum f_i x_i = 40k + 1910$

The mean is given as $21$. Using the formula for the mean $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$:

$\dfrac{40k + 1910}{2k + 90} = 21$

$40k + 1910 = 21(2k + 90)$

$40k + 1910 = 42k + 1890$

$2k = 20 \Rightarrow k = 10$

Now, substituting $x = 10$ into the given options to check which equation has $10$ as a root:

For $2x^2 - 19x - 10 = 0$:
$2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$

Thus, $k = 10$ is a root of $2x^2 - 19x - 10 = 0$.

Answer: $2x^2 - 19x - 10 = 0$