To solve the equation √2(x+1) = 4x-5, we need to isolate the variable x.

First, let's square both sides of the equation to eliminate the square root:

(√2(x+1))^2 = (4x-5)^2

2(x+1) = (4x-5)(4x-5)

Expanding the right side:

2(x+1) = 16x^2 - 40x - 40x + 25

2(x+1) = 16x^2 - 80x + 25

Distribute the 2 on the left side:

2x + 2 = 16x^2 - 80x + 25

Rearrange the equation to set it equal to zero:

16x^2 - 82x + 23 = 0

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. However, this equation does not factor easily, so we will use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 16, b = -82, and c = 23. Substituting these values into the quadratic formula:

x = (-(-82) ± √((-82)^2 - 4(16)(23))) / (2(16))

Simplifying:

x = (82 ± √(6724 - 1472)) / 32

x = (82 ± √5252) / 32

x = (82 ± 72.47) / 32

Now, we have two possible solutions:

x = (82 + 72.47) / 32 ≈ 4.82

x = (82 - 72.47) / 32 ≈ 0.31

Therefore, the solutions to the equation √2(x+1) = 4x-5 are approximately x = 4.82 and x = 0.31.