If you are sure that's exactly what the equation is:
b,c,d are all numbers, so indeed, focus on a single number, say 'e'. Also a can't be zero, or there is no equation at all! So, again all numbers can be grouped:
a*sqrt(x) + b+c=d ---> sqrt(x) = f, f some number.
x = f^2 is the solution.
To be extraneous, means that it does not solve the equation.
Why here? Because by definition in mathematics sqrt( f^2) is not f! but abs(f), that is:
sqrt(9)=3,
So you just need to set the numbers to -3: (thinking x=9)
2*sqrt(x) + 7 -4 = -3
this gives: sqrt(x) = -3, from which x= 9, but sqrt(9) = 3.
It's kind of tricky, because many forget the definition that sqrt(a^2) = abs(a).
The same does not happen when looking for roots.
You should ask your teacher to make sure this is what they want.
Let me say that usually extraneous solutions are those which come out after solving an equation but make no sense when substituted in the original equation. But with sqrt(x) and your so particular equation, there is nothing else to say.