jmc

We expand both sides: $$\begin{array}{rl}(4x+11)(2x-8)& =x(2x+7)\\ 8x^2-10x-88& =2x^2+7x\\ 6x^2-17x-88& =0\\ (2x-11)(3x+8)& =0\end{array}$$Therefore, the smaller of the two solutions is $x=-8/3$ and the larger is $x=\boxed{\frac{11}{2}}$