# How do you find the zeros, real and imaginary, of y=x^2 +36x +81 using the quadratic formula?

Apr 20, 2016

Zeros are $- 18 - 9 \sqrt{3}$ and $- 18 + 9 \sqrt{3}$

#### Explanation:

Zeros of $y = {x}^{2} + 36 x + 81$ are given by solution to the equation

${x}^{2} + 36 x + 81 = 0$

Hence using quadratic formula $x$ is given by $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Hence $x = \frac{- 36 \pm \sqrt{{36}^{2} - 4 \times 1 \times 81}}{2} = \frac{- 36 \pm \sqrt{1296 - 324}}{2}$

= $\frac{- 36 \pm \sqrt{972}}{2} = \frac{- 36 \pm \sqrt{3 \times 3 \times 3 \times 3 \times 3 \times 2 \times 2}}{2} = \frac{- 36 \pm 18 \sqrt{3}}{2}$

= $- 18 \pm 9 \sqrt{3}$

Hence zeros are $- 18 - 9 \sqrt{3}$ and $- 18 + 9 \sqrt{3}$