Need help with one of these: - b + √ b² - 4ac

Can someone please explain I did the problem twice and still got it wrong: Here is the problem
z² -2z -11 =0

I got
2 ± √-4 - 4 (1)(-11)
----------------------------
2(1)

= 2 ± √40
---------------
2

= 2 ± √2³(5)
------------------
2

= 2 ± 2√10
------------------
2

= 1 ± 2√5

What did I do wrong, can someone show me how to do this?

... z² - 2z -11 = 0
The Quadratic Formula:
z = [ -b ± √ ( (b)² - 4(a)(c) ) ] / 2(a)
z = [ -(-2) ± √ ( (-2)² - 4(1)(-11) ) ] / 2(1)
z = [ 2 ± √ ( 4 + 44 ) ] / 2
z = [ 2 ± √48 ] / 2
z = [ 2 ± √(16*3) ] / 2
z = [ 2 ± 4√3 ] / 2
z = 1 ± 2√3

Formula: {(-b)+-sqrt(b^2 - 4ac)]/2a
= {-(-2)+-sqrt[(-2)^2 - 4(1)(-11)]}/2(1)
= {2+-sqrt[4 +44]}/2
= {2+-sqrt(48)}/2
= {2+-4sqrt(3)}/2
= 1+-2sqrt(3)

(2 ± 2√(10))/2 = 1 ± √(10)

√(10) = √(2)√(5), NOT 2 √(5)

In any case, this is incorrect since (–2)² = +4

You got
2 ± √-4 - 4 (1)(-11) ///// should be √[4 - 4 (1)(-11)] = √[4+44] = √48
----------------------------
2(1)

= 2 ± √48
---------------
2

= 2 ± 4√3
------------------
2

= 1 +/- 2 √3

a=1, b=-2,c=-11

2+or- sqrt4-4(1)(-11) 2+or-sqrt48 2+or-4sqrt3
----------------------------- = ----------- = ---------------- = 1+or-2sqrt3

2(1) 2 2

-0

in your first line b^2 should be 4 not -4 since -2^2 = 4