Log2(x-5)+log2(x+2)=3

ОДЗ
$\left \{ {{x-5\ \textgreater \ 0} \atop {x+2\ \textgreater \ 0}} \right. \\ \\ \left \{ {{x\ \textgreater \ 5} \atop {x\ \textgreater \ -2}} \right. \\ \\ x\ \textgreater \ 5$

$log_2(x-5)+log_2(x+2)=3 \\ \\ log_2(x-5)(x+2)=3 \\ \\ (x+2)(x-5)=2^3 \\ \\ x^{2} -5x+2x-10=8 \\ \\ x^{2} -3x-18=0 \\ \\ D=3^2+4*18= 81 \\ \\ x_1= \frac{3+9}{2}=6 \\ \\ x_2=\frac{3-9}{2}=-3$

х₂=-3 не удовлетворяет ОДЗ

Ответ х=6