1+log7(x+4)=log7(x2+9x+20)

$1+log_7(x+4)=log_7(x^2+9x+20)$
ОДЗ: $\left \{ {{x+4\ \textgreater \ 0} \atop {x^2+9x+20\ \textgreater \ 0}} \right. ; \left \{ {{x\ \textgreater \ -4} \atop {(x+4)(x+5)\ \textgreater \ 0}} \right. ;\left \{ {{x\ \textgreater \ -4} \atop {x+4\ \textgreater \ 0;x+5\ \textgreater \ 0}} \right. ;\left \{ {{x\ \textgreater \ -4} \atop {x\ \textgreater \ -4;x\ \textgreater \ -5}} \right. ;x\ \textgreater \ -4$
$log_7 7(x+4)=log_7(x^2+9x+20)$
$7(x+4)=x^2+9x+20$
$7x+28-x^2-9x-20=0$
$-x^2-2x+8=0$

$(x+4)(2-x)=0$
$x_1=-4;x_2=2$
$x=2$