(4x+3)^2\2=72 and x > 0 then x =?

(4x + 3)^2 / 2 = 36     multiply both sides by 2

(4x + 3)^2  = 72         take the square root of both sides

4x + 3 = ±√72            subtract 3 from both sides

4x = ±√72 - 3                divide by 4 on both sides

x = [ ±√72 - 3] / 4

And since you specify that x > 0, then the solution is x = [ √72 - 3] / 4 = [ 6√2 - 3] / 4