First of all, we have to set the equation in the form [tex]p(x)=0[/tex], where [tex]p(x)[/tex] is a quadratic polynomial. So, we have
[tex]3x^2-5x=0[/tex]
Even though using the quadratic formula for such a simple case is a little bit overkill, it is surely doable: we have
[tex]3x^2-5x=0 \iff ax^2+bx+c=0 \iff a=3,\quad b=-5,\quad c=0[/tex]
And the formula for solving is
[tex]x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
So, if we plug our values, we have
[tex]x_{1,2} = \dfrac{5\pm\sqrt{25}}{6} = \dfrac{5\pm5}{6}[/tex]
And thus the solutions are
[tex]x_1=0,\quad x_2 = \dfrac{10}{6}=\dfrac{5}{3}[/tex]