$x=\frac{2}{3} \to (\frac{2}{3})^2+\frac{2}{3}b-4=0 \to \frac{4}{9}+\frac{2}{3}b-4=0  \to \frac{2}{3}b=\frac{32}{9}\to b=\frac{16}{3}$ $x^2+\frac{16}{3}x-4=0$ $\Delta=(\frac{16}{3})^2-4(1)(-4)=\frac{16^2}{9}+16=\frac{400}{9} \to \sqrt {\Delta}=\pm \frac{20}{3}$ $x_{1,2}=\frac{-\frac{16}{3}\pm \frac{20}{3}}{2} \to x_1=\frac{2}{3 }, x_2=-6$