Question 39964
You are correct...you have to get the variable y onto one side of the equation.
But you should use parentheses to clarify the problem;

Is it sqrt(2y+7) + 4 = y ?
or is it sqrt(2y) + 7 + 4 = y ?
or perhaps sqrt(2y+ 7 + 4) = y ?

I am going to assume that it's:

{{{sqrt(2y+7) + 4 = y}}} Subtract 4 from both sides of the equation.
{{{sqrt(2y+7) = y - 4}}} Now square both sides.
{{{2y+7 = (y-4)^2}}} Simplify.
{{{2y+7 = y^2 - 8y + 16}}} Subtract 2y from both sides.
{{{7 = y^2 - 10y + 16}}} Now subtract 7 from both sides.
{{{y^2 - 10y + 9 = 0}}} Solve by factoring.
{{{(y - 1)(y - 9) = 0}}} Apply the zero products principle.
{{{y - 1 = 0}}} and/or {{{y - 9 = 0}}}
If {{{y - 1 = 0}}}, then {{{y = 1}}}
If {{{y - 9 = 0}}}, then {{{y = 9}}}

Now if youcheck these solutions by substituting y=1 and then y=9 into your original equation, you will find that y=9 works if you use only the positive value of {{{sqrt(25)}}}and y=1 works if you use only the negative value of {{{sqrt(9)}}}.

{{{sqrt(2y+7) + 4 = y}}} Try y = 3.
{{{sqrt(2(9)+7) + 4 = 9}}}
{{{sqrt(18+7) + 4 = 9}}}
{{{sqrt(25) + 4 = 9}}}
{{{5 + 4 = 9}}} Only the positive value  of {{{sqrt(25)}}} works.

{{{sqrt(2y+7) + 4 = y}}} Try y = 1.
{{{sqrt(2(1)+7) + 4 = 1}}}
{{{sqrt(2+7) + 4 = 1}}}
{{{sqrt(9) + 4 = 1}}}
{{{-3 + 4 = 1}}} Only the negative value of {{{sqrt(9)}}} works.