f(x)=3x2−12x+12f(x) = 3x^{2} - 12x + 12f(x)=3x2−12x+12 and f(x)=3f(x) = 3f(x)=3
∴f(x)=3=3x2−12x+12 ⟹ 3x2−12x+12−3=0\therefore f(x) = 3 = 3x^{2} - 12x + 12 \implies 3x^{2} - 12x + 12 - 3 = 0∴f(x)=3=3x2−12x+12⟹3x2−12x+12−3=0
3x2−12x+9=0;3x2−9x−3x+9=03x^{2} - 12x + 9 = 0; 3x^{2} - 9x - 3x + 9 = 03x2−12x+9=0;3x2−9x−3x+9=0
3x(x−3)−3(x−3)=(3x−3)(x−3)=03x(x - 3) - 3(x - 3) = (3x - 3)(x - 3) = 03x(x−3)−3(x−3)=(3x−3)(x−3)=0
3x - 3 = 0 or x - 3 = 0
x=1 or 3x = 1 \text{ or } 3x=1 or 3
