Complete the square...?

(x+2)^2 -4-1=0
(x+2)^2 -5=0
(x+2)^2 = 5

0 = x^2 + 4x - 1
x^2 + 2x + 2x - 1 = 0
x^2 + 2x + 2x = 1
x^2 + 2x + 2x + 4 = 1 + 4
(x^2 + 2x) + (2x + 4) = 5
x(x + 2) + 2(x + 2) = 5
(x + 2)(x + 2) = 5
(x + 2)^2 = 5

(x + 2)^2 = 5
x + 2 = ±√5
x = -2 ±√5

∴ x = -2 ±√5

first you want a symmetrical quadratic that fills the equation as much as possible:
(x+2)^2
This would expand out to give you:
(x^2)+4x+4
You now need to get this to the original by removing 5 as x^2 + 4x + 4 - 5 gives your original - so your completed square is:
(x+2)^2 - 5

From here you can work out the value of x:
0 = (x+2)^2 - 5
5 = (x+2)^2
root(5) = x+2
root(5) -2 = x
therefore:
x = 0.236067977
or approx:
x = 0.24

( x^2 + 4x + 4 ) - 5 = 0

( x + 2 )^2 = 5

x + 2 = ± √5

x = - 2 ± √5

Add 5 to both sides:

0 + 5 = (x^2 + 4x - 1 + 5)
5 = (x^2 + 4x + 4)
5 = (x + 2)^2
+/- sqrt(5) = x + 2

x = -2 +/- sqrt ( 5 )