let's first off, recall that the factoring of the quadratic, will come from the factoring of the constant, whose FOIL factors added will give the middle term.
now, let's firstly multiply both sides by the LCD of 3, to do away with the denominator.
[tex]\bf x^2-3t=\cfrac{v}{3}x\implies \stackrel{\textit{multiplying both sides by 3}}{3x^2-9t=vx}\implies 3x^2-vx-9t=0[/tex]
now, since our constant of -9, can only be factored as 3*3 OR 9*1, so
[tex]\bf 3x^2-vx-9t=0\implies \begin{array}{llll} (3x\pm 3)(x\pm 3)=0\\\\ (3x\pm 1)(x\pm 9)=0 \end{array} \\\\[-0.35em] ~\dotfill\\\\ 3x\pm 3=0\implies 3x=\mp 3\implies x=\mp\cfrac{3}{3}\implies \boxed{x=\mp 1} \\\\[-0.35em] ~\dotfill\\\\ x\pm 3=0\implies \boxed{x=\mp 3} \\\\[-0.35em] ~\dotfill\\\\ 3x\pm 1=0\implies 3x=\mp 1\implies \boxed{x=\mp \cfrac{1}{3}} \\\\[-0.35em] ~\dotfill\\\\ x\pm 9 = 0\implies \boxed{x=\mp 9}[/tex]
now, we could have also used the rational root test for that, and get the same variations.
