SOLVE THE EQUATION squareroot (x-5)-4=3?

I can't tell where the radical starts and ends but let's suppose it covers all of the left side. Then, square both sides and get

x-5-4 = 9

x - 9 = 9

x = 18

or if the left expression has the sqrt of only the first expression

add 4 to each side

sqrt(x-5) = 7

square both sides

x -5 = 49

x = 54

sqrt(x-5)-4+4=3+4

sqrt(x-5)=7

(sqrt(x-5))^2=7^2

x-5=49

x-5+5=49+5

x=54

remember to isolate the radical then square both sides (or cube for third roots, etc.)

also remember if a=b then a^c=b^c

check

sqrt(54-5) - 4 = 3

sqrt(49) - 4 = 3

7- 4 = 3

3 = 3

sqrt(x - 5) - 4 = 3

Add 4 to both sides:

sqrt(x - 5) = 7

Square both sides:

x - 5 = 49

Add 5 to both sides:

x = 54

(x-5)-4=3

x-5-4=3

x-9=3

x-9+9=3+9

x=12