Given,
3(x−2)2=147⇒3(x2−4x+4)=147⇒3x2−12x+12=147⇒3x2−12x+12−147=0 (Writing as ax2+bx+c=0)⇒3x2−12x−135=0⇒33x2−12x−135=30 (Dividing the complete equation by 3) ⇒x2−4x−45=0⇒x2−9x+5x−45=0⇒x(x−9)+5(x−9)=0⇒(x−9)(x+5)=0 (Factorising left side) ⇒x−9=0 or x+5=0 (Zero-product rule) ⇒x=9 or x=−5.
Hence, the roots of given equation are 9, -5.