Let's try solving this question using a substitution. We begin by wishing that the pesky #-1# wasn't on the left hand side of the equation, otherwise we'd just square both sides to get rid of the square roots! So let's substitute something that would allow us to do this, let
#v=sqrt(x)-1#
Then solving for #x# we get
#x=(v+1)^2 = v^2+2v+1#
We can now substitute the first expression for #v# into the left hand side of the equation, and the second one under the square-root on the right hand side:
#sqrt(v^2+2v-6)=v#
now we can square both sides:
#v^2+2v-6=v^2#
We notice the the #v^2#'s on either side cancel out, and then we can solve for #v# giving:
#2v-6 = 0 implies v=3#
Now we can use our expression for #x# above to solve for #x#
#x=(v+1)^2 = 4^2 = 16#
