Question 624279
{{{sqrt(5x-15)-sqrt(x+1)=2}}} , adding {{{sqrt(x+1)}}} to both sides, turns into
{{{sqrt(5x-15)=sqrt(x+1)+2}}}
Squaring both sides, we get {{{5x-15=x+1+4*sqrt(x+1)+4}}}
Adding {{{-x-5}}} to both sides we get
{{{4x-20=4*sqrt(x+1)}}}
Dividing both sides by 4 we get
{{{x-5=sqrt(x+1)}}}
Suqaring both sides again, we get
{{{x^2-10x+25=x+1}}}
Adding {{{-x-1}}} to both sides, we get
{{{x^2-11x+24=0}}}
Factoring, we get
{{{(x-3)(x-8)=0}}}
The solutions to {{{x^2-11x+24=0}}} are {{{x=3}}} and {{{x=8}}}.
Those could be solutions of the original equation, or could be extraneous solutions, introduced by all that squaring of both sides.
We must test them in the original equation.
{{{x=3}}} gives {{{sqrt(5x-15)=0}}} while {{{sqrt(x+1)=sqrt(4)=2}}},
so {{{sqrt(5x-15)-sqrt(x+1)=-2}}} 
showing that{{{x=3}}} is not a solution of the original equation.
{{{x=8}}} gives {{{sqrt(5x-15)=sqrt(40-15)=sqrt(25)=5}}} while {{{sqrt(x+1)=sqrt(9)=3}}},
so {{{sqrt(5x-15)-sqrt(x+1)=5-3=2}}} 
showing that {{{highlight(x=8)}}} is the solution.