Solve the following equations by using formula:

$\dfrac{3x - 4}{7} + \dfrac{7}{3x - 4} = \dfrac{5}{2}, x$ ≠ $\dfrac{4}{3}$

The given equation is $\dfrac{3x - 4}{7} + \dfrac{7}{3x - 4} = \dfrac{5}{2}$

$\Rightarrow \dfrac{(3x - 4)^2 + 7^2}{7(3x - 4)} = \dfrac{5}{2} \text{ (On taking L.C.M. ) }\\[1em] \Rightarrow \dfrac{9x^2 + 16 - 24x + 49}{21x - 28} = \dfrac{5}{2} \\[1em] \Rightarrow 2(9x^2 - 24x + 65) = 5(21x - 28) \\[1em] \Rightarrow 18x^2 - 48x + 130 = 105x - 140 \\[1em] \Rightarrow 18x^2 - 48x - 105x + 130 + 140 = 0 \\[1em] \Rightarrow 18x^2 - 153x + 270 = 0 \\[1em] \Rightarrow 9(2x^2 - 17x + 30) = 0 \\[1em] \Rightarrow 2x^2 - 17x + 30 = 0$

Comparing it with ax² + bx + c = 0
a= 2, b = -17, c = 30

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-17) ± \sqrt{(-17)^2 - 4 \times 2 \times 30}}{2 \times 2} \\[1em] \Rightarrow \dfrac{17 ± \sqrt{289 - 240}}{4} \\[1em] \Rightarrow \dfrac{17 + \sqrt{49}}{4} \text{ or } \dfrac{17 - \sqrt{49}}{4} \\[1em] \Rightarrow \dfrac{17 + 7}{4} \text{ or } \dfrac{17 - 7}{4} \\[1em] \Rightarrow \dfrac{24}{4} \text{ or } \dfrac{10}{4} \\[1em] 6 \text{ or } \dfrac{5}{2}$

Hence, roots of the given equation are $6, \dfrac{5}{2}$.