\[x² - bx + 2b - 3 = 0\]
\[D = b^{2} - 8b + 12 > 0\]
\[b^{2} - 8b + 12 = 0\]
\[b_{1} + b_{2} = 8,\ \ b_{1} \cdot b_{2} = 12\]
\[b_{1} = 6,\ \ b_{2} = 2\]
\[Ответ:( - \infty;2) \cup (6;\ + \infty).\]