parametric equations help

The curve with parametric equations $\displaystyle x = 3t^{2} , y = 4t^{3} $ along with the tangent at the point with parameter t = 1

find the equation if this tangent and the coordinates of the point where it meets the curve again.

$\displaystyle \frac{dx}{dt} = 6t $

$\displaystyle \frac{dy}{dt} = 12t^{2} $

$\displaystyle \frac{dy}{dx} = \frac{12t^{2}}{6t} = 2t $

when t = 1

$\displaystyle x = 3 , y = 4 , \frac{dy}{dx} = 2 $

equation of tangent = $\displaystyle y -4 = 2(x-3) $

$\displaystyle y = 2x-2 $

however to find the coordinates of where the curve crosses the tangent again, would I not just be able to sub in the values of x and y into the tnagent equation?

$\displaystyle 4t^{3} = 2(3t^{2})-2 $

$\displaystyle 4t^{3} = 6t^{2} - 2 $ can't seem to solve from here,

any help appreicated.

thanks