Answer:
[tex]x_{1} =\frac{-17- \sqrt{129}}{2}, x_{2} =\frac{-17+ \sqrt{129}}{2} [/tex]
Step-by-step explanation:
[tex] \frac{x + 2}{x + 3} + \frac{x - 3}{x + 2} = \frac{5}{2} [/tex]
[tex] \frac{x + 2}{x + 3} + \frac{x - 3}{x + 2} = \frac{5}{2} [/tex]
[tex] \frac{x + 2}{x + 3} + \frac{x - 3}{x + 2} - \frac{5}{2} = 0[/tex]
[tex] \frac{2(x + 2 {)}^{2} + 2(x + 3) \times (x - 3) - 5(x + 3) \times (x + 2) }{2(x + 3) \times (x + 2)} = 0[/tex]
[tex] \frac{2(x + 2 {)}^{2} + 2( {x}^{2} - 9) + ( - 15x - 15) \times (x + 2) }{2(x + 3) \times (x + 2)} = 0 [/tex]
[tex] \frac{2( {x}^{2} + 4x + 4) - 3 {x}^{2} - 48 - 25x}{2(x + 3) \times (x + 2)} = 0[/tex]
[tex] \frac{2 {x}^{2} + 8x + 8 - 3 {x}^{2} - 48 - 25x }{2(x + 3) \times (x + 2)} = 0[/tex]
[tex] \frac{ - {x}^{2} - 17x - 40 }{2(x + 3) \times (x + 2)} = 0[/tex]
[tex] {x}^{2} + 17x + 40 = 0[/tex]
[tex]x = \frac{ - 17± \sqrt{ {17}^{2} - 4 \times 1 \times 40 } }{2 \times 1} [/tex]
[tex]x = \frac{ - 17± \sqrt{289 - 160} }{2} [/tex]
[tex]x = \frac{ - 17± \sqrt{129} }{2} [/tex]
[tex]x_{1} =\frac{-17- \sqrt{129}}{2}, x_{2} =\frac{-17+ \sqrt{129}}{2} [/tex]
