# How do you find the zeros, real and imaginary, of y=x^2-31x-9 using the quadratic formula?

${x}_{1} = \frac{+ 31 + \sqrt{\left(961 + 36\right)}}{2} = 31.2877$
and ${x}_{2} = \frac{+ 31 - \sqrt{\left(961 + 36\right)}}{2} = - 0.287653$

#### Explanation:

from the given: $y = {x}^{2} - 31 x - 9$

$a = 1$, $b = - 31$ , $c = - 9$

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

$x = \frac{- - 31 \pm \sqrt{{\left(- 31\right)}^{2} - 4 \left(1\right) \left(- 9\right)}}{2 \cdot 1}$

$x = \frac{+ 31 \pm \sqrt{\left(961 + 36\right)}}{2}$

${x}_{1} = \frac{+ 31 + \sqrt{\left(961 + 36\right)}}{2} = 31.2877$
and ${x}_{2} = \frac{+ 31 - \sqrt{\left(961 + 36\right)}}{2} = - 0.287653$

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