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

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Answer 1 · 2017-10-23T23:37:10

Roots are $= \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$ $= ( 1 + isqrt3)/2, ( 1 - isqrt3)/2$

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

Roots are $\frac{- b \pm \sqrt{b^{2} - (4 \cdot a \cdot c)}}{2 a}$ $(-b +- sqrt(b^2 - (4*a*c)))/(2a)$

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

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

Roots are $\frac{1 \pm \sqrt{1 - 4}}{2}$ $( 1 +- sqrt(1 - 4))/2$

Roots are $= (\frac{1}{2}) (1 + i \sqrt{3}), (\frac{1}{2}) (1 - i \sqrt{3})$ $= (1/2)( 1+ isqrt3), (1/2)( 1- isqrt3)$

How do you find the zeros, real and imaginary, of y=x2−x+1y=x^2 -x+1 using the quadratic formula?