Answer
no real solutions
Work Step by Step
In the given equation, \begin{align*} 3x^2-4x+3=0 ,\end{align*} $a= 3 ,$ $b= -4 ,$ and $c= 3 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{align*}\require{cancel}x&= \dfrac{-(-4)\pm\sqrt{(-4)^2-4(3)(3)}}{2(3)} \\\\&= \dfrac{4\pm\sqrt{16-36}}{6} \\\\&= \dfrac{4\pm\sqrt{-20}}{6} \\\\&=
\dfrac{4\pm\sqrt{4\cdot-5}}{6}
\\\\&=
\dfrac{4\pm\sqrt{(2)^2\cdot-5}}{6}
\\\\&=
\dfrac{4\pm2\sqrt{-5}}{6}
.\end{align*}
Since the radicand of the square root symbol is negative, then there are no real solutions.
