Решите уравнение: (2x^2+x-3)/(2x+3)=0.

\[\frac{2x^{2} + x - 3}{2x + 3} = 0\ \ \ \ \ \ \ \ \ \ \ x \neq - 15\ \]

\[2x^{2} + x - 3 = 0\]

\[D = 1^{2} - 4 \cdot 2 \bullet ( - 3) = 1 + 24 =\]

\[= 25\]

\[x_{1} = \frac{- 1 + \sqrt{25}}{2 \cdot 2} = \frac{- 1 + 5}{4} = \frac{4}{4} =\]

\[= 1\]

\[x_{2} = \frac{- 1 - \sqrt{25}}{2 \cdot 2} = \frac{- 1 - 5}{4} =\]

\[= \frac{- 6}{4} = - \frac{3}{2} =\]

\[= - 1,5\ \ (не\ подходит)\]

\[Ответ:1.\]