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

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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#