X-x^2=1, x equals with

Set this equal to 0:

-x^2 + x - 1 = 0

Check the value of the discriminant to check the number of solutions for this quadratic problem:

D = b^2 -4ac = 1 - 4(-1)(-1) = 1 - 4 = -3

Since the discriminant is negative, there is no real x value to satisfy this problem

Hope this helps

Dear Nichole,

x-x^(2)=1

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
-x^(2)+x-1=0

Divide each term in the equation by -1.
x^(2)-x+1=0

Use the quadratic formula to find the solutions. In this case, the values are a=1, b=-1, and c=1.
x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

Substitute in the values of a=1, b=-1, and c=1.
x=(-(-1)\~((-1)^(2)-4(1)(1)))/(2(1))

Multiply -1 by each term inside the parentheses.
x=(1\~((-1)^(2)-4(1)(1)))/(2(1))

Simplify the section inside the radical.
x=(1\i~(3))/(2(1))

Simplify the denominator of the quadratic formula.
x=(1\i~(3))/(2)

Simplify the expression to solve for the + portion of the \.
x=(1+i~(3))/(2)

Simplify the expression to solve for the - portion of the \.
x=(1-i~(3))/(2)

The final answer is the combination of both solutions.
x=(1+i~(3))/(2),(1-i~(3))/(2)

x-x^2=1
transfer x^2 and x to the other side, you will get :
x^2-x+1=0
now use the quadratic formula to get,
x = (1±sqrt(3)*i)/2 , where i is the imaginary number ( square root of -1).

Hope this helps! :)

well it can't be factorised unless you use the complex numbers
factorising it with complex numbers would give you :
0.5 plus -0.86603 i
0.5 minus -0.86603 i

x^2-x+1=0

a=1 b=-1 c=1

1+/- sq rt -1^2-4(1)(1)/2
x=1+/- sq rt 1-4/2
x=1+/- sq rt -3/2
x=1+/-i sq rt 3/2