We have the next quadratic function given:
[tex]3x^2-9x+2=0[/tex]Mary used the next quadratic formula:
[tex]x=\frac{-b^\pm\sqrt{b^2-4ac}}{2a}[/tex]Replace using the form ax²+bx+c
Where a= 3
b=-9
c=2
Then:
[tex]\begin{gathered} x=\frac{-\lparen-9)\pm\sqrt{\left(-9\right)^2-4\left(3\right)\left(2\right)}}{2\left(3\right)} \\ x=\frac{9\pm\sqrt[]{57}}{6}\frac{}{} \end{gathered}[/tex]Therefore, the correct answer is "x = fraction numerator 9 plus-or-minus square root of 57 over denominator 6 end fraction"