Use quadratic formula to solve:

$\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$

Given,

$\Rightarrow \dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{x - 7 - x - 4}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{-11}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow -11 \times 30 = 11(x + 4)(x - 7) \\[1em] \Rightarrow \dfrac{-11 \times 30}{11} = x^2 - 7x + 4x - 28 \\[1em] \Rightarrow -30 = x^2 - 3x - 28 \\[1em] \Rightarrow x^2 - 3x + 2 = 0$

Now we have,

x² - 3x + 2 = 0

Comparing x² - 3x + 2 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -3 and c = 2.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \dfrac{3 \pm \sqrt{1}}{2} \\[1em] = \dfrac{3 \pm 1}{2} \\[1em] = \dfrac{3 + 1}{2} \text{ or } \dfrac{3 - 1}{2} \\[1em] = \dfrac{4}{2} \text{ or } \dfrac{2}{2} \\[1em] = 2 \text{ or } 1.$

Hence, x = 1 or 2.