Question

How do you find the zeros, real and imaginary, of `y=x^2 -x+1` using the quadratic formula?

Answer 1

Roots are `= ( 1 + isqrt3)/2, ( 1 - isqrt3)/2`

Explanation:

Standard form of quadratic equation is
`ax^2 + bx + c`

Roots are `(-b +- sqrt(b^2 - (4*a*c)))/(2a)`

`y = x^2 -x - 1`

`a = 1, b = -1, c = 1`

Roots are `( 1 +- sqrt(1 - 4))/2`

Roots are `= (1/2)( 1+ isqrt3), (1/2)( 1- isqrt3)`