Опишите решение, будьте добры.

Решите задачу:

$\sqrt{x+5}-\sqrt{2x+3}=1, \\ \left \{ {{x+5 \geq 0,} \atop {2x+3 \geq 0;}} \right. \left \{ {{x \geq -5,} \atop {x \geq -1,5;}} \right. \ x \geq -1,5, \\ \sqrt{x+5}=\sqrt{2x+3}+1, \\ (\sqrt{x+5})^2=(\sqrt{2x+3}+1)^2, \\ x+5=2x+3+2\sqrt{2x+3}+1, \\ 2\sqrt{2x+3})=1-x, \\ 1-x \geq 0, \ x \leq 1, \\ (2\sqrt{2x+3})^2=(1-x)^2, \\ 4(2x+3)=1-2x+x^2, \\ x^2-10x-11=0, \\ x_1=-1, \ x_2=11\ \textgreater \ 1, \\ x=-1.$