Алгебра: Корень(x+20)-корень(x-1)=3

Решите задачу:

$\sqrt{x+20}-\sqrt{x-1}=3\; ,\; \; ODZ:\; \; \left \{ {{x+20\geq 0} \atop {x-1\geq 0}} \right. \; \left \{ {{x\geq -20} \atop {x\geq 1}} \right. \; \Rightarrow \; \; x\geq 1\\\\\sqrt{x+20}=3+\sqrt{x-1}\\\\x+20=9+2\cdot 3\cdot \sqrt{x-1}+x-1\\\\6\sqrt{x-1}=12\\\\\sqrt{x-1}=2\\\\x-1=4\\\\x=5\\\\Proverka:\; \; \sqrt{5+20}-\sqrt{5-1}=\sqrt{25}-\sqrt4=5-2=3\; ;\; \; 3=3\\\\Otvet:\; \; x=5.$