Answer:
No real solution
Step-by-step explanation:
[tex]3x^2-4x+4=0\quad\implies\quad a=3\,,\quad b=-4\,,\quad c=4\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot3\cdot4}}{2\cdot3}=\dfrac{4\pm\sqrt{-32}}{6}[/tex]
There is no solution in real number as it's no real √[-32]
{solutins in complex numbers would be: [tex]x_1=\frac{4+4\sqrt{2}\,i}{6}=\frac{2+2\sqrt{2}\,i}{3}\\ x_2=\frac{2-2\sqrt{2}\,i}{3}[/tex] }
