Hallo Adriano,
der Rechenweg geht so:$$\begin{aligned} \frac{x-8}{x+5} &=\frac{x+4}{x-7} &&|\,x\ne -5 \land x \ne 7\\& &&|\, \cdot(x+5)(x-7) \\ (x-8)(x-7) &= (x+4)(x+5) \\ x^2 - 15x + 56 &= x^2 + 9x + 20 &&|\,-x^2 \\ - 15x + 56 &= + 9x + 20 &&|\,-56 \\ - 15x &= + 9x -36 &&|\,-9x \\ - 24x &= -36 &&|\,\div (-24) \\ x &= \frac{-36}{-24} \\ x &= \frac 32 \\ \end{aligned}$$Gruß Werner