There are two ways to do this problem.  The way you're probably
studying now is u-substitution:

       x-7√x+10 = 0

Let u = √x, then u² = x

       u²-7u+10 = 0

     (u-2)(u-5) = 0

  u-2 = 0;  u-5 = 0
    u = 2     u = 5

   √x = 2;     √x = 5 
 
(√x)² = 2²; (√x)² = 5²
    x = 4;      x = 25

We must always check radical
equations to find out if either
or both of those answers are 
legitimate solutions or are
only extraneous.

Checking x = 4:

       x-7√x+10 = 0
       4-7√4+10 = 0
       4-7·2+10 = 0
        4-14+10 = 0
              0 = 0

So 4 is a legitimate solution.

Checking x = 25:

       x-7√x+10 = 0
      25-7√25+10 = 0
      25-7·5+10 = 0
       25-35+10 = 0
              0 = 0 

So 25 is also a legitimate solution.

Edwin

Here's the other way to solve that problem:

     x-√x+10 = 0

        x+10 = 7√x

     (x+10)² = (7√x)²

(x+10)(x+10) = (7√x)²

  x²+20x+100 = 49x
  x²-29x+100 = 0
 (x-25)(x-4) = 0

  x-25 = 0;  x-4 = 0
     x = 25;   x = 4

Checking is the the same as in the other way to solve it.

Edwin

1. Isolate that square root term on one side of the equation.
2. Square both sides