solution to x^2-2x-2=0?

Answer 1

Using the quadratic formula,
x=(2+ sqrt{(-2)^2+8})/2 or x=(2- sqrt{(-2)^2+8})/2

Therefore, x =1+sqrt{3} or x =1- sqrt{3}

Answer 2

X 2 2x 2 0

Answer 3

X2 2x 2 0

Answer 4

x = 1 + 3^1/2 and x = 1 - 3^1/2

Answer 5

because factoring over the set of integers does not work, try completing the square method.
add 2 to both sides, so you have:
x^2-2x=2
complete the square on the left side by adding 1, but you add 1 to the right side as well:
x^2-2x+1=2+1
(x-1)^2=3
square root of both sides:
x-1=+/- sqr 3
solutions are:
x= 1+sqr3 or x=1-sqr3

Answer 6

x²-2x-2=0
it is an equ of second degree (we look to the hygher power of x in this line witch is 2)
its general form is

ax²+bx+c=0
with a=1(because 1*x²=x²),b=-2,c=-2

we have to calculate
Δ=(b)²-4(a*b)=(-2)²-4(-2*1)
=4+8=12
((Squared root of Δ=12^1/2=(4*3)^1/2
=((2²)(3))^1/2=2*(3^1/2).
((=3.46))
Δ is > 0 (12>0) our equ accepte two solutions

X1=
-b+Sqrt ofΔ
--------------
2a

X2=
-b-Sqrt ofΔ
----------------
2

then,
X1=
-(-2)+2*(3^1/2)
-----------------------
2
=
2+2*(3^1/2)
-------------------
2
=
X1=1+3^1/2=2.73
and,
X2=1-3^1/2= -0.73

Answer 7

This equation is unfactorable, we can use the quadratic formula.
when a=1 b=-2 c=-2
x=(-b+/-sqrt[b^2-4ac])/2a

x=(-(-2)+/-sqrt[(-2)^2-4(1)(-2)]/2(1)
x=(2+/-sqrt[4+8])/2
x=(2+/-sqrt[12])/2
x=(2+/sqrt[4*3])/2
x=(2+/-2sqrt[3])/2

there for x1= (2+2sqrt[3])/2 or 1+sqrt(3) and x2=(2-2sqrt[3])/2 or 1-sqrt(3)

Answer 8

It's not factorable, so you need to use the quadratic formula.

Do you know the song? Well heres the song to remember the quadratic formula: X equals the opposite of b, plus or minus the square root of b squared minus 4ac all divided by 2 a.

It uses the pop up jack in the box song.

Then you subsitute everything into the formula.

You end up with...

1 ± root 3

Answer 9

You can also complete the square. I get x = 1 plus or minus sqrt(3).

Answer 10

u'll have to use the quadratic formula or complete square to solve this problem. It can't be factored out.