How do i solve x^2+2=x?

x²+2 = x
x²-x+2 = 0

Complete the square:
x²-x+(1/4)+(7/4) = 0
(x - 1/2)² + (√7/2)² = 0
(x - 1/2 - (√7/2)i)(x - 1/2 + (√7/2)i) = 0
x = {1/2 ± (√7/2)i}

You can also use the quadratic formula to get the same answer.

x²-x+2 = 0
ax²+bx+c=0
x = [-b±√(b²-4ac)]/(2a)

a = 1
b = -1
c = 2

x = [1±√((-1)²-4(1)(2))]/2
= [1±√(-7)]/2
= (1±i√7)/2

which agrees with the answer obtained by completing the square

Question Number 1 :
For this equation x^2 + 2 = x , answer the following questions :
A. Find the roots using Quadratic Formula !
B. Use completing the square to find the root of the equation !

Answer Number 1 :
First, we have to turn equation : x^2 + 2 = x , into a*x^2+b*x+c=0 form.
x^2 + 2 = x , move everything in the right hand side, to the left hand side of the equation
<=> x^2 + 2 - ( x ) = 0 , which is the same with
<=> x^2 + 2 + ( - x ) =0 , now open the bracket and we get
<=> x^2 - x + 2 = 0

The equation x^2 - x + 2 = 0 is already in a*x^2+b*x+c=0 form.
In that form, we can easily derive that the value of a = 1, b = -1, c = 2.

1A. Find the roots using Quadratic Formula !
Use the formula,
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
We had know that a = 1, b = -1 and c = 2,
we need to subtitute a,b,c in the abc formula, with thos values.
Which produce x1 = (-(-1) + sqrt( (-1)^2 - 4 * (1)*(2)))/(2*1) and x2 = (-(-1) - sqrt( (-1)^2 - 4 * (1)*(2)))/(2*1)
Which can be turned into x1 = ( 1 + sqrt( 1-8))/(2) and x2 = ( 1 - sqrt( 1-8))/(2)
Which is the same with x1 = ( 1 + sqrt( -7))/(2) and x2 = ( 1 - sqrt( -7))/(2)
Which can be turned into x1 = ( 1 + sqrt(7)*sqrt(-1))/(2) and x2 = ( 1 - sqrt(7)*sqrt(-1))/(2)
Because we know that sqrt(-1) = i,
It imply that x1 = ( 1 + 2.64575131106459*i )/(2) and x2 = ( 1 - 2.64575131106459*i )/(2)
So we have the answers x1 = 0.5 + 1.3228756555323*i and x2 = 0.5 - 1.3228756555323*i

1B. Use completing the square to find the root of the equation !
x^2 - x + 2 = 0 ,divide both side with 1
So we get x^2 - x + 2 = 0 ,
We know that the coefficient of x is -1
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -1/2 = -0.5
By using that fact we turn the equation into x^2 - x + 0.25 + 1.75 = 0
Which can be turned into ( x - 0.5 )^2 + 1.75 = 0
Which is the same with (( x - 0.5 ) - 1.3228756555323*i ) * (( x - 0.5 ) + 1.3228756555323*i ) = 0
By opening the brackets we will get ( x - 0.5 - 1.3228756555323*i ) * ( x - 0.5 + 1.3228756555323*i ) = 0
We get following answers x1 = 0.5 - 1.3228756555323*i and x2 = 0.5 + 1.3228756555323*i

here given that
x²-x+2 = 0
use the formula if
ax²+bx+c=0 then
x = [-b±√(b²-4ac)]/(2a)..............(1)
now in our case
a = 1,b = -1,c = 2
plug or put the values in ---(1)
x = [-{(-1)±√(-1)²-4(1)(2)}]/2
= [1±√(-7)]/2
= (1±i√7)/2 where i=√(-1)
Answer is roots are imaginery

someone forgot to reverse the sign

X^2-x+2=0

y = ax^2 + bx +c
a=1: b=1: c=2
for y=0:

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

b^2=1

4ac=8

sqrt(1-8) is negative so no real roots to this equation

x^2+2=x
x^2-x+2 = 0 its a quadratic equation in x where
a = x^2 coefficient = 1
b = x coefficient = -1
c = constant = 2

discriminat = b^2 - 4ac

=(-1)^2 - 4 ( 1) (2)
=1-8
= -7 which is less than zero
so the roots of quadratic equation are imaginary.

x^2 + 2 = x
x^2 - x + 2 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -1
c = 2

x = [1 ±√(1 - 8)]/2
x = [1 ±√-7]/2 (imaginary number)
(no real roots)

x²+2 = x
x²-x+2 = 0

Complete the square:
x²-x+(1/4)+(7/4) = 0
(x - 1/2)² + (√7/2)² = 0
(x - 1/2 - (√7/2)i)(x - 1/2 + (√7/2)i) = 0
x = {1/2 ± (√7/2)i}

X^2+2=x
X^2-x+2=0
D=1-8<0 => It has no solutions

you shift the x to the other side, which will make the equation x^2-x+2=0. Now you find the two values for x by either factorising the equation or using the quadratic formula.