**Question:** Find the value of $\displaystyle x$ when $\displaystyle y = 0$
$\displaystyle y = \frac{2x - 1}{x + 3}$

**Attempt:**
$\displaystyle (2x - 1)(x + 3)^{-1} = 0$

$\displaystyle {\color{red}(2x - 1)(x^{-1} + \frac{1}{3}) = 0}$

$\displaystyle 2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0$

$\displaystyle \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0$

I think the quadratic equation that i got is wrong ... Where did I go wrong?