Решите пожалуйста:

Решите задачу:

$5^x = t\\ t \ \textgreater \ 0\\\\ \frac{8}{t-3} - \frac{6}{t+1} = 3\\\\ \frac{8(t+1)-6(t-3)}{(t-3)(t+1)}=3\\\\ \frac{8t+8-6t+18}{(t-3)(t+1)}=3\\\\ \frac{2t+26}{(t-3)(t+1)}=3\\\\ 3(t-3)(t+1)=2t+26\\ 3(t^2+t-3t-3)-2t-26=0\\ 3t^2+3t-9t-9-2t-26=0\\ 3t^2-8t-35=0\\ D = 64 + 4*3*35=484, \sqrt{D} = 22\\\\ t_1=\frac{8-22}{6} = -\frac{14}{6}, \varnothing\\\\ t_2=\frac{8+22}{6} = 5\\\\ 5^x=5^1\\ x=1\\$