Solve the following equation by factorisation:

$\dfrac{1}{7}(3x-5)^2 = 28$

Given,

$\dfrac{1}{7}(3x-5)^2 = 28 \\[1em] \Rightarrow \dfrac{1}{7}( 9x^2 - 30x + 25) = 28 \\[1em] \Rightarrow \dfrac{9}{7}x^2 - \dfrac{30}{7}x + \dfrac{25}{7} = 28 \\[1em] \Rightarrow \dfrac{9}{7}x^2 \times 7 - \dfrac{30}{7}x \times 7 + \dfrac{25}{7} \times 7 = 28 \times 7 \text{ (Multiplying the complete equation by 7) } \\[1em] \Rightarrow 9x^2 - 30x + 25 = 196 \\[1em] \Rightarrow 9x^2 - 30x + 25 - 196 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 9x^2 - 30x - 171 = 0 \\[1em] \Rightarrow 9x^2 - 57x + 27x - 171 = 0 \\[1em] \Rightarrow 3x(3x - 19) + 9(3x - 19) = 0 \\[1em] \Rightarrow (3x + 9)(3x - 19) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 3x + 9 = 0 \text{ or } 3x - 19 = 0 \text{ (Zero - product rule) }\\[1em] \Rightarrow 3x = -9 \text { or } 3x = 19. \\[1em] \Rightarrow x = -\dfrac{9}{3} \text{ or } x = \dfrac{19}{3} \\[1em] \Rightarrow x = -3 \text{ or } x = \dfrac{19}{3} \\[1em]$

Hence, the roots of given equation are -3, $\dfrac{19}{3}$.