Answer
$\{2+\frac{\sqrt 3}{3}, 2-\frac{\sqrt 3}{3}\}$
Work Step by Step
$3x^{2} -12x+11 = 0$
$3x^{2} -12x = -11$
Dividing both sides by $3$
$x^{2} -4x = \frac{-11}{3}$
Adding the square of half the co-efficient of $x$ to both sides,
$x^{2} -4x +4= \frac{-11}{3}+4$
$x^{2} -4x +4= \frac{-11+12}{3}$
$x^{2} -4x +4= \frac{1}{3}$
$(x-2)^{2}= \frac{1}{3}$
$(x-2)= ±\sqrt \frac{1}{3}$
$(x-2)= ±\frac{1}{\sqrt 3}$
Rationalizing the denominator,
$(x-2)= ±\frac{1}{\sqrt 3} \times \frac{\sqrt 3}{\sqrt 3}$
$(x-2)= ± \frac{\sqrt 3}{ 3}$
$x= 2± \frac{\sqrt 3}{ 3}$
$x= 2+\frac{\sqrt 3}{ 3}$ or $x= 2- \frac{\sqrt 3}{ 3}$
