\[(b + 1)x^{2} + (b + 3)x + 2 = 0;\ \ \]
\[b + 1 = 0 \Longrightarrow b = - 1\]
\[D = (b + 3)^{2} - 4 \cdot (b + 1) \cdot 2 =\]
\[= b^{2} + 6b + 9 - 8b - 8 =\]
\[= b^{2} - 2b + 1 = (b - 1)^{2}\]
\[(b - 1)^{2} = 0\]
\[b - 1 = 0\]
\[b = 1.\]
\[Ответ:при\ b = 1;\ b = - 1.\]