how do i solve this equation with the quadratic formula?

Question

i need help solving this quadratic formula

x^2-11=2x

x^2-2x-11 <-- standard form

i think it is -2+4sq.root3/2 but im not sure
helpp? =)

Answer 1

You're on the right track, x^2-2x-11 is the standard form.

So we have a = 1, b = -2, c = -11.

Now you only need to put these numbers into the quadratic formula:

x_1 / x_2 = (-(-2) +- sqrt ((-2)²- 4 *1 * (-11))) / 2*1 = (2 +- sqrt (4 + 44)) / 2 = (2 +- sqrt (48))/2 = (2 +- sqrt (16*3))/2 = (2 +- 4 * sqrt (3))/2 = 1+- 2 sqrt (3)

So the first solution is 1- 2 sqrt (3) and the second solution is 1+ 2 sqrt (3)

Answer 2

First, standard form is that your expression=0.
You can actually solve this a little faster by noting that x^2-2x+1 = (x-1)^2. Then we can write
x^2-2x+1-12 = 0 and (x-1)^2 = +/-sqrt(12)= +/-2 sqrt(3). From this you get
x = -1-2sqrt(3) or x =-1+2sqrt(3)
Somehow you have doubled your answer. Were you dividing by 2 in the quadratic formula?

Answer 3

nicely the formulation for fixing an quadratic equation is -b + - the sq. root of b squared + 4 x A rewrite the equation like this 8x + x -5 8x=A x=b and -5 is c -x + and minus x to the 2nd ability + 4 x 8 x -5 be squared equals 8x8 it is sixty 4 + 4 x a x -5 a good 20 x X 20 x a million is 20 the sq. root of 20 is 4.40 seven + -8 = -3.fifty 3 then minus is is 11.fifty 3 so which you have 2 solutions 11.fifty 3 constructive and -3.fifty 3 unfavourable variety thats how i do quadratic equations I dont be attentive to if its ideal that's advisable to examine with somebody who is familiar with extra relating to the stuff like a math instructor

Answer 4

x^2 -2x -11 = 0

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

using the quadratic formula

- (-2) =/- ((4-4(1)9-11))^1/2 / 2

2 +/- (4 + 44)1/2 /2

2 +/- (48) 1/2 /2

2 +/- (16 x 3)^1/2 /2

2 +/- 4(3)^1/2 / 2 cancel the 2's

1 +/- 2(3)^1/2

x = 1 + 2(3)^1/2

x = 1 - 2(3)^1/2

Answer 5

x= [ 2 +/- sqrt( 2^2+4*11) ] /2
= [2 +/- sqrt(48) ]/2
= [2 +/- 4 sqrt(3) ]/2
= 1 + 2sqrt(3) or 1 - 2sqrt(3)

if you complete the square
x^2 - 2x - 11 = (x-1)^2 - 12
and its easier to see that x = +1 +/- 2sqrt(3)

Answer 6

Well, your origianl EQ, in st. form is ax^2 + bx + c = 0

So the Quadratic would then be:

x= [ -b +/- (SQRT(b^2 - 4 a c)) ] / 2a

Now just sub in the coef. and solve for x

Answer 7

your answer is nearly right.
but there will be two roots:
1+(2sq.root3) & 1-(2sq.root3) [since the 2 in denominator will get cancelled with the common 2 in the numerator.]

Answer 8

x^2-1=2x subtract 2x from each side
x^2-2x-11=0

x=(2+/-√(4+11*4))/2
x=(2+/-√(48))/2
x=1+/-2√3
x-1+2√3
x=1-2√3

Answer 9

x²-2x-11

a = 1
b = -2
c = -11

the quadratic formula is
x = [ -b +- sqrt ( b² - 4(a)(c)) ] / 2a

substitute in your values and you get

2 +- sqrt(48)

remember, its "all over 2a", i think you messed up there

Answer 10

x=1+2sqrt(3) and 1-2sqrt(3)