help please

√(x + 7) - √(6 - x)   = 1        add    √(6 - x)  to both sides to separate the radicals

√(x + 7)  =  1 + √(6 - x)      square both sides

x + 7  = 1 + 2 √(6 - x)  + (6 -x)    rearrange to get theradical on one side

2x  =2 √(6 - x)      divide both sides by 2

x  = √(6 - x)      square both sides, again

x^2  = 6 - x      rearrange

x^2 + x  - 6 = 0      factor

(x + 3) ( x - 2) = 0

And setting each factor to 0, we have that x = -3  or x = 2 ......we must check these in the original problem, as these types of equations often lead to "extraneous" solutions [due to  the squaring of both sides]

Check -3    ... [√(-3 + 7) - √(6 - (-3)) = 1 ] =   [2 - 3   =  1 ]  is untrue

Check 2  ...  [ √(2 + 7) - √(6 - 2) = 1 ]    =  [ 3 - 2 ] = 1 which is true

So x = 2 is the only solution.......here's a graph of both sides  of the original problem.....

https://www.desmos.com/calculator/rmqhq2arba

Notice that the only solution occurs at (2, 1)