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

Oct 23, 2017

Roots are $= \frac{1 + i \sqrt{3}}{2} , \frac{1 - i \sqrt{3}}{2}$

#### Explanation:

Standard form of quadratic equation is
$a {x}^{2} + b x + c$

Roots are $\frac{- b \pm \sqrt{{b}^{2} - \left(4 \cdot a \cdot c\right)}}{2 a}$

$y = {x}^{2} - x - 1$

$a = 1 , b = - 1 , c = 1$

Roots are $\frac{1 \pm \sqrt{1 - 4}}{2}$

Roots are $= \left(\frac{1}{2}\right) \left(1 + i \sqrt{3}\right) , \left(\frac{1}{2}\right) \left(1 - i \sqrt{3}\right)$