(½)x2 + 2x + 1 = 0 please help!!?

Question

can someone please please help me solve this equation... I have been trying to solve it for at least half an hour!!

(½)x2 + 2x + 1 = 0

-2±√2
2±√2
2±√3

Answer 1

(½)x2 + 2x + 1 = 0
multiply both sides by2
x^2 +4x+2=0

x = [-b+-sq root of (b^2-4ac] /2a

=[-4 +-sq root of (16-8)] /2

= (-4±2√2 ) /2

= -2±√2

Answer 2

½)x2 + 2x + 1 = 0

It's not going to factor. So I fall back on the Quadratic Formula

You could multiply both sides by 2 to get rid of the fraction, but it really doesn't make any difference.
a= 1/2
b = 2
c = 1

The Quadratic Formula
x = (-(b) ±√[(b)² - 4(a)(c)])/[2(a)]
Substituting
x = {-(2) ±√[(2)² - 4(1/2)(1)]} / [2(1/2)]

Doing some of the arithmetic...
x = (-2 ±√[4 - 2])/1

And a little more...
x = -2 ±√[2]

Answer 3

Use the quadratic formula with
a = 1/2, b = 2, c = 1
thus x = [-2 +/- sqrt(4-4*1/2*1)] / (2*(1/2))
x = [-2 +/- sqrt(2)]/1
x = -2 +/ sqrt(2)

The first choice is the answer.

Answer 4

Use the quadratic formula:
(-b±√(b^2-4ac)) / (2a)
-2 ±√ (4-2) / 1

-2±√2

Answer 5

x² + 4x + 2 = 0
x = [ - 4 ± √(16 - 8) ] / 2
x = [ - 4 ± √8 ] / 2
x = [- 4 ± 2√2] / 2
x = - 2 ± √2
1st OPTION

Answer 6

well it has to be the first one becasue the formula starts off as -b hence the first digit should be a negative...... no calculation needed, think its just common sense ... hope it is right though