First we need to write down this equation
[tex] \sqrt{x+3} + 4 = 6[/tex]
now we switch 4 to other side and subtract it from 6
[tex] \sqrt{x+3} =2[/tex]
Now we square both sides
x+3 = 4
and we get x = 1
If we replace x=1 in starting equation we get 6=6 which is not extraneus
Answer is first option
