√3x+4 - √2x+4 = 2

Isolate either one of the radicals.
I'll isolate the first radical on the left side:

            √3x+4 = 2 + √2x+4

Square both sides:

         (√3x+4)² = (2 + √2x+4)²

             3x+4 = (2 + √2x+4)(2 + √2x+4)

FOIL out the right side

             3x+4 = 4 + 2√2x+4 + 2√2x+4 + 2x+4

             3x+4 = 4 + 4√2x+4 + 2x+4

             3x+4 = 8 + 4√2x+4 + 2x

Isolate the remaining radical term on the right:

              x-4 = 4√2x+4

Square both sides:

           (x-4)² = (4√2x+4)²

       (x-4)(x-4) = 16(2x+4)

         x²-8x+16 = 32x+64

        x²-40x-48 = 0

That doesn't factor so we have to use the quadratic formula:

                x =  

                x =  

                x =   

                x = 
     
                x = 

                x = 

                x = 

                x = 4(5 ± 2√7)

Using the +,    x = 4(5 + 2√7), approximately 41.16601049

Using the -,    x = 4(5 - 2√7), approximately -1.166010489

We must check for extraneous solutions:

    √3x+4 - √2x+4 = 2

Checking 41.2  (rounded to tenths)

√3(41.2)+4 - √2(41.2)+4 = 2
√127.6 - √86.4 = 2
11.296 - 9.295 = 2
         2.001 = 2

Not a perfect check, but close enough to believe it's 
a solution, sibce we rounded off.

x = 4(5 + 2√7), 

Checking -1.17  (rounded to hundredths)

√3(-1.17)+4 - √2(-1.17)+4 = 2
√2(41.2)+4 - √.49 = 2
   .7 - 1.66 = 2
        -.96 = 2

That's not close at all, so we believe it
is extraneous, and not a solution.

So there is just one solution:  x = 4(5 + 2√7).

Edwin