How do you find the zeros, if any, of $y = - 7 x^{2} - x + 22$ $y= -7x^2-x+22$ using the quadratic formula?

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Answer 1 · 2018-05-23T12:46:35

$x = - \frac{1 + \sqrt{617}}{14} or - \frac{1 - \sqrt{617}}{14}$ $x=-(1+sqrt(617))/14or-(1-sqrt(617))/14$

$x = - 1.846 or 1.702$ $x=-1.846 or 1.702$

Given $a x^{2} + b x + c$ $ax^2+bx+c$ , the zeros are given by:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$a = - 7$ $a=-7$
$b = - 1$ $b=-1$
$c = 22$ $c=22$

$x = \frac{1 \pm \sqrt{{(- 1)}^{2} - 4 (- 7) (22)}}{2 (- 7)}$ $x=(1+-sqrt((-1)^2-4(-7)(22)))/(2(-7))$

$x = \frac{1 \pm \sqrt{1 + 616}}{- 14}$ $x=(1+-sqrt(1+616))/-14$

$x = - \frac{1 \pm \sqrt{617}}{14}$ $x=-(1+-sqrt(617))/14$

$x = - \frac{1 + \sqrt{617}}{14} or - \frac{1 - \sqrt{617}}{14}$ $x=-(1+sqrt(617))/14or-(1-sqrt(617))/14$

$x = - 1.846 or 1.702$ $x=-1.846 or 1.702$

How do you find the zeros, if any, of y= -7x^2-x+22y=−7x2−x+22y= -7x^2-x+22using the quadratic formula?