Question:

The curve $\displaystyle y^2 = 12x$ intersects the line $\displaystyle 3y = 4x + 6$ at two points. Find the distance between the two points.

Attempt:

$\displaystyle y^2 = 12x$

$\displaystyle y = (12x)^{\frac{1}{2}}$

$\displaystyle 3y = 4x + 6$

$\displaystyle y = \frac{4}{3}x + 2$

$\displaystyle (12x)^{\frac{1}{2}} = \frac{4}{3}x + 2$

$\displaystyle 12x = \left(\frac{4}{3}x + 2\right)^2$

$\displaystyle 12x = \left(\frac{4}{3}x + 2\right)\left(\frac{4}{3}x + 2\right)$

$\displaystyle 12x = \frac{16}{9}^2 + \frac{16}{3}x + 4$

$\displaystyle = \frac{16}{9}^2 + \frac{16}{3}x - 12x + 4$

$\displaystyle = \frac{16}{9}^2 - \frac{20}{3}x + 4$

$\displaystyle a = \frac{16}{9} \ , \ b = -\frac{20}{3} \ , \ c = 4$

$\displaystyle x = \frac{-b \pm \sqrt(b^2 - 4ac)}{2a}$

$\displaystyle x = \frac{-(-\frac{20}{3}) \pm \sqrt((-\frac{20}{3})^2 - 4 \times \frac{16}{9} \times 4)}{2 \times \frac{16}{9}}$

Where did I go wrong?