Question 78525
{{{sqrt((2*y)+7)+4=y}}}
First, get the terms under the radical on one side of the equation:
{{{sqrt((2*y)+7)=y-4}}}
Now, square both sides of the equation:
{{{(2*y)+7=(y-4)^2}}}
Simplify, and put the equation into {{{ax^2+bx+c=0}}} form:
{{{(2*y)+7=y^2-(8*y)+16}}}
{{{y^2-(10*y)+9=0}}}
Factor this expression as follows:
{{{(y-1)*(y-9)=0}}}
Note that this is a second order equation, so there are two roots.
Equate each of the factors you just found to zero and solve for y.
You get:
{{{y=1}}} and {{{y=9}}}

Now, this part is important... you need to plug these answers
back in to the original problem to be sure they are not "extraneous."
An extraneous root is mathematically corrrect, but not a true answer.

If you plug y=1 back into the original equation, you will see that the
equation does NOT hold. Hence this is an extraneous root.

If you plug y=9 back into the original equation, you will see that the
equation DOES hold. so that one is your answer.