wait never mind I got it

Start: sqrt(11 - x)   -   sqrt(x + 6)  = 3

Move sqrt(x + 6) to other side: sqrt(11 - x) = 3 + sqrt(x + 6)

Square both sides: [sqrt(11 - x)]^2 = (3 + sqrt(x + 6))^2  =>  11 - x = (3 + sqrt(x + 6))^2

Expand RHS: 11 - x = 3^2 + 2*3*sqrt(x + 6) + sqrt(x + 6)^2

Simplify: 11 - x = 9 + 6 * sqrt(x + 6) + (x+6)

More: 11 - x = x + 15 + 6 * sqrt(x + 6)

More: 11 - x - x - 15 = 6 * sqrt(x + 6)

More: -4 - 2x = 6 * sqrt(x + 6)

Square both sides: (-4 - 2x)^2 = (6 * sqrt(x + 6))^2

Expand: (-4)^2 + -4*-2x + -2x*-4 + (-2x)^2 = 36 * (x+6)

Simplify: 16 + 2(8x) + 4x^2 = 36(x+6)

More: 16 +16x + 4x^2 = 36x + 216

More: 16-216 + 16x-36x + 4x^2 = 0

More: -200 -20x +4x^2 = 0

Divide by 4: x^2 - 5x - 50 = 0

Its a nice quadratic now!

2^(x +1) = 7^(x + 2)

Pulling out the known powers: 2^x * 2 = 7^x * 49

Then: 2/49 = 7^x / 2^x

Then: 2/49 = (7/2)^x

Using that special property of logarithms I cannot remember name of: log(2/49) = log(7/2) x

Rearranging: x = log(2/49)/log(7/2)

Calculator: x = -2.55


Or maybe neater this way:

2^(x +1) = 7^(x + 2)

Using that special property of logarithms I cannot remember name of: (x+1) log 2 = (x+2) log 7

expanding: x log 2 + log 2 = x log 7 + 2 log 7

pulling x's to one side: x log 2 - x log 7 = 2 log 7 - log 2

Then : x (log 2 - log 7) = 2 log 7 - log 2

Then: x = (2 log 7 - log 2) / (log 2 - log 7)

Calculator: x = 1.389 / -0.544 = -2.55


... test:  2^(x+1) = 7^(x+2)  ==>  2^(-2.55+1) = 7^(-2.55+2)  ==>  2^(-1.55) = 7^(-0.55)  ==>  0.341 = 0.343 (close enough)