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

Question

1586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-23T12:46:35

$x=-(1+sqrt(617))/14or-(1-sqrt(617))/14$

$x=-1.846 or 1.702$

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

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

$x=(1+-sqrt((-1)^2-4(-7)(22)))/(2(-7))$

$x=(1+-sqrt(1+616))/-14$

$x=-(1+-sqrt(617))/14$

$x=-(1+sqrt(617))/14or-(1-sqrt(617))/14$

$x=-1.846 or 1.702$

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