Find the roots of the following equation:
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq -4, 7$

(A) Given equation: $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
Taking the common denominator on the left side:
$\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
Simplifying the numerator:
$\frac{x-7-x-4}{x^2-7x+4x-28}=\frac{11}{30}$
$\frac{-11}{x^2-3x-28}=\frac{11}{30}$
Dividing both sides by $11$:
$\frac{-1}{x^2-3x-28}=\frac{1}{30}$
Cross-multiplying:
$x^2-3x-28 = -30$
$x^2-3x-28+30 = 0$
$x^2-3x+2 = 0$
Factoring the quadratic equation:
$x^2-2x-x+2 = 0$
$x(x-2)-1(x-2) = 0$
$(x-2)(x-1) = 0$
Therefore,the roots are $x = 1$ or $x = 2$.