Solve the following equation by factorisation:

3(x - 2)² = 147

Given,

$3(x-2)^2 = 147 \\[1em] \Rightarrow 3( x^2 - 4x + 4) = 147 \\[1em] \Rightarrow 3x^2 - 12x + 12 = 147 \\[1em] \Rightarrow 3x^2 - 12x + 12 - 147 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 3x^2 - 12x - 135 = 0 \\[1em] \Rightarrow \dfrac{3x^2 - 12x - 135 }{3} = \dfrac{0}{3} \\[1em] \text{ (Dividing the complete equation by 3) } \\[1em] \Rightarrow x^2 - 4x - 45 = 0 \\[1em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[1em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[1em] \Rightarrow (x - 9)(x + 5) = 0 \text{ (Factorising left side) }\\[1em] \Rightarrow x - 9 = 0 \text{ or } x + 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 9 \text { or } x = -5.$

Hence, the roots of given equation are 9, -5.