help on this algebra equation?

x² - 2x - 8 = 0

x = [- b ± √ (b² - 4ac) ] / 2a

x = [ 2 ± √ (4 + 32) ] / 2

x = [ 2 ± 6 ] / 2

x = 4 , - 2

2x = x² - 8

x² - 2x - 8 = 0

quadratic formula:

x = [-(-2) ± √((-2)² - 4(1)(-8))]/(2(1))

 = [2 ± √36]/2

 = [2 ± 6]/2

 = 4, -2

2x = x^2 - 8

x^2 - 2x - 8 = 0

x

= [-b +/- (b^2 - 4ac)^(1/2)] / 2a

= [2 +/- (4 + 32)^(1/2)] / 2

= 1 +/- 3

= -2 or 4

x² – 2x – 8 = 0

quadratic equation:

to solve ax² + bx + c = 0

x = [–b ± √(b²–4ac)] / 2a

x = [2 ± √(4+32)] / 2

x = [2 ± √(36)] / 2

x = [2 ± 6)] / 2

x = 1 ± 3

x = 4, –2

0 = x^2 - 2x - 8.

The "a,b,c" of this quadratic are 1, -2, and -8.

Therefore

x = -b/2a +/- (1/2a)*sqrt(b^2 - 4ac)

= 1 +/- (1/2)*sqrt(4+32)

= 1 +/- 3;

the solutions are +4 and -2.

To use quadratic equation, you need your equation in standard form:

0 = ax² + bx + c

So to do that, subtract 2x from both sides and put in proper order:

2x = x² - 8

0 = x² - 2x - 8

Now substitute the following values into the equation and simplify:

a = 1

b = -2

c = -8

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -(-2) ± √((-2)² - 4(1)(-8))] / (2 * 1)

x = [ 2 ± √(4 + 32)] / 2

x = [ 2 ± √(36)] / 2

x = (2 ± 6) / 2

x = -4/2 and 8/2

x = -2 and 4