# How do you find the zeros, real and imaginary, of y=x^2-5x+8 using the quadratic formula?

Apr 27, 2018

$x = \frac{5}{2} + \frac{\sqrt{7}}{2} i$ and $x = \frac{5}{2} - \frac{\sqrt{7}}{2} i$

#### Explanation:

We have:

$y = {x}^{2} - 5 x + 8$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

We get the the roots of the equation $y = 0$:

$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(8\right)}}{2 \left(1\right)}$

$\setminus \setminus = \frac{5 \pm \sqrt{25 - 32}}{2}$

$\setminus \setminus = \frac{5 \pm \sqrt{- 7}}{2}$

$\setminus \setminus = \frac{5 \pm \sqrt{7} i}{2}$

Leading to the two conjugate roots:

$x = \frac{5}{2} + \frac{\sqrt{7}}{2} i$ and $x = \frac{5}{2} - \frac{\sqrt{7}}{2} i$