What is the solution to the equation 1 over the square root of 8 = 4(m + 2)

Solve for m:
1/sqrt(8) = 4 (m + 2)

Rationalize the denominator. 1/sqrt(8) = 1/sqrt(8)×(8^(1 - 1/2))/(8^(1 - 1/2)) = (8^(1 - 1/2))/8:
(8^(1 - 1/2))/8 = 4 (m + 2)

Combine powers. (8^(1 - 1/2))/8 = 8^((1 - 1/2) - 1):
8^((1 - 1/2) - 1) = 4 (m + 2)

Put 1 - 1/2 over the common denominator 2. 1 - 1/2 = 2/2 - 1/2:
8^(2/2 - 1/2 - 1) = 4 (m + 2)

2/2 - 1/2 = (2 - 1)/2:
8^((2 - 1)/2 - 1) = 4 (m + 2)

2 - 1 = 1:
8^(1/2 - 1) = 4 (m + 2)

Put 1/2 - 1 over the common denominator 2. 1/2 - 1 = 1/2 - 2/2:
8^(1/2 - 2/2) = 4 (m + 2)

1/2 - 2/2 = (1 - 2)/2:
8^((1 - 2)/2) = 4 (m + 2)

1 - 2 = -1:
8^((-1)/2) = 4 (m + 2)

1/sqrt(8) = 1/sqrt(2^3) = (1/sqrt(2))/2:
(1/sqrt(2))/2 = 4 (m + 2)

Rationalize the denominator. 1/(2 sqrt(2)) = 1/(2 sqrt(2))×(sqrt(2))/(sqrt(2)) = (sqrt(2))/(2×2):(sqrt(2))/(2×2) = 4 (m + 2)

2×2 = 4:
(sqrt(2))/4 = 4 (m + 2)

(sqrt(2))/4 = 4 (m + 2) is equivalent to 4 (m + 2) = (sqrt(2))/4:
4 (m + 2) = sqrt(2)/4

Expand out terms of the left hand side:
4 m + 8 = sqrt(2)/4

Subtract 8 from both sides:
4 m + (8 - 8) = (sqrt(2))/4 - 8

8 - 8 = 0:
4 m = (sqrt(2))/4 - 8
Put each term in (sqrt(2))/4 - 8 over the common denominator 4: (sqrt(2))/4 - 8 = (sqrt(2))/4 - 32/4:
4 m = (sqrt(2))/4 - 32/4

(sqrt(2))/4 - 32/4 = (sqrt(2) - 32)/4:
4 m = (sqrt(2) - 32)/4

Divide both sides by 4:
  m = (sqrt(2) - 32)/(4×4)

4should be the last one n=4