What is the greatest possible value of $x$ that solves the equation \[ \left( \frac {4x - 16}{3x - 4} \right)^{\!2} + \left( \frac {4x - 16}

$u=\dfrac{4x-16}{3x-4}\\ u^2 + u -12=0\\ (u+4)(u-3) = 0\\ u=-4,3\\ \dfrac{4x-16}{3x-4}=-4 \Rightarrow x=2\\ \dfrac{4x-16}{3x-4}=3 \Rightarrow x=-\dfrac 4 5$

$2 \text{ is the greatest possible solution}$