quadriatc formula?

Question

idk if i am doing this right can u tell me ... heres the problem
formula =ax(squared)=bx=c=0

x=-b+-(square root)b(squared)-4ac
-------------------------------------------------
2a
x2 (squared)-2x-10=0

for my answer i got 2+-2Square root 11 over 2

Answer 1

x^2 - 2x - 10 = 0
x = [-(-2) +/- sqrt{(-2)^2 - 4*1*(-10)}]/2
x = [2 +/- sqrt(4 + 40)]/2
x = [2 +/- sqrt(44)]/2
x = [2 +/- sqrt(4)*sqrt(11)]/2
x = [2 +/- 2*sqrt(11)]/2
x = 1 +/- sqrt(11)

Answer 2

Quadratic Formula: x = [-b +/- V`b^2 - 4ac] / 2a

Equation: x^2 - 2x - 10 = 0

First: identify three terms for three variables...
a = 1; b = -2; c = -10

Sec: substitute the terms with the corresponding variables.

x = [-(-2) +/- V`(-2)^2 - 4(1)(-10)] / 2(1)

x = [2 +/- V`(-2)(-2) - 4(1)(-10)] / 2

x = [2 +/- V`4 - 4(1)(-10)] / 2

x = [2 +/- V`4 + 40] / 2

x = [2 +/- V`44] / 2

x = [2 +/- V`2*2*11] / 2

x = [2 +/- 2 V`11] / 2

Third: you will have 2 solutions.

a. x = [2 + 2 V`11] / 2
x = 2/2 + [2 V`11]/2
x = 1 + V`11

b. x = [2 - 2 V`11] / 2
x = 2/2 - [2 V`11]/2
x = 1 - V`11

Solutions: 1 + V`11 and 1 - V`11

Answer 3

x² - 2x - 10 = 0
x = [2 ± √44] / 2
x = [2 ± 2√11] / 2 (which is your answer)
x = 1 ± √11

Answer 4

The quadratic formula is x=(-b±√(b^2-4ac))/(2a)

x^2-2x-10
x=(2±√(4+40))/2
x=(2±√44)/2
x=1±√11

You got it right except for canceling 2's in the numerator & denominator.

Answer 5

Yes, your right. However, you could simplify your result to get
1 +- sqrt(11)
"one plus or minus square root of eleven"


How come?
Just like this (let's say "v" means "square root"):
2+-2v11
-------------
2

Then, we can see there's a common factor of 2:

2(1 +- v11)
----------------
2

Now we can simplify this 2 to get the final answer:

1+- v11

So the first root equals 1 + v11 and the 2nd root equals 1 - v11.
Right?

Answer 6

You got the right answer.