How do you find the zeros of # y = -x^2 - 12x + 7 # using the quadratic formula?

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Answer 1 · 2015-11-28T07:40:58

The zeros are at
#x = -6+sqrt(43)# and #x = -6 - sqrt(43)#

The quadratic formula states that
#ax^2 + bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)#

We are looking for the values of #x# at which

#-x^2 - 12x + 7 = 0#

Then we can apply the quadratic formula with
#a=-1, b=-12, c=7#
to obtain

#x = (-(-12)+-sqrt((-12)^2-4(-1)(7)))/(2(-1))#

#= (12+-sqrt(144 + 28))/(-2)#

#= -(12+-sqrt(172))/2#

#= -(12 +- 2sqrt(43))/2#

#= -6 +- sqrt(43)#

Thus the zeros are at
#x = -6+sqrt(43)# and #x = -6 - sqrt(43)#