Question 172937
if you have variable inside a square root you have to do the opposit operation :squaring to "take out" x from root  
[example {{{(sqrt(3))^2=(3 ^(1/2))^2=3^((1/2)*2)=3^1=3}}}]

{{{sqrt(2x+5) -sqrt(x-2)=3}}}  square both sides of the equation
{{{(sqrt(2x+5))^2 -(sqrt(x-2))^2=3^2}}} 
2x+5-(x-2)=9
x=2  answer