Finding asymptotes for $f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$ $$f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$$
I know that the horizontal asymptote is $1/3$. To find the vertical asymptotes, I set the denominator equal to zero and used the quadratic formula, and I got $-5$ and $2/3$, and this is wrong. 
How do you find the vertical asymptotes for this problem? 
thanks. 

$3x^2 + 13x - 10 = (3x - 2)(x + 5)$, and $x^2 + 3x - 10 = (x + 5)(x - 3)$. Thus only $x = \dfrac{2}{3}$ is the vertical asymptote since $\displaystyle \lim_{x \to -5} f(x) = \dfrac{8}{17} \neq \infty$
