Question 116235
 Solve algebraically and check your potential solutions show work
squareroot x+20-x=0
:
Without the use of brackets, it's rather ambiguous what is inside the radical;
I am assuming:
{{{sqrt(x+20)}}} - x = 0
:
Add x to both sides:
{{{sqrt(x+20)}}} = x
:
Square both sides:
x + 20 = x^2
:
Arrange as a quadratic on the right:
0 = x^2 - x - 20
:
This will factor to:
(x-5)(x+4) = 0
Solutions
x = +5 
and
x = -4
:
Check solutions in original equation:
x=5
{{{sqrt(5+20)}}} - 5 = 0
{{{sqrt(25)}}} - 5 = 0
5 - 5 = 0
x=-4
{{{sqrt(-4+20)}}} -(-4) = 0
x=-4
{{{sqrt(16)}}} + 4) = 0
4 + 4 does not = 0, not a solution