# How do you find the zeros, real and imaginary, of y=x^2 -18x +81 using the quadratic formula?

Nov 20, 2015

Only one solution $x = 9$
As $\left(\text{Re"+"Im}\right) \to \left(9 + 0 i\right)$

#### Explanation:

Given: $y = {x}^{2} - 18 x + 81$

Using std form of $y = a {x}^{2} + b x + c$

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

$a = 1$
$b = \left(- 18\right)$
$c = 81$

Thus

$x = \frac{- \left(- 18\right) \pm \sqrt{{\left(- 18\right)}^{2} - 4 \left(1\right) \left(81\right)}}{2 \left(1\right)}$

$x = \frac{18 \pm \sqrt{324 - 324}}{2}$

$x = \frac{18}{2} = 9$