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

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Answer 1 · 2016-02-09T07:14:57

#x_1=(+31+sqrt((961+36)))/(2)=31.2877#
and #x_2=(+31-sqrt((961+36)))/(2)=-0.287653#

from the given: #y=x^2-31x-9#

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

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(--31+-sqrt((-31)^2-4(1)(-9)))/(2*1)#

#x=(+31+-sqrt((961+36)))/(2)#

#x_1=(+31+sqrt((961+36)))/(2)=31.2877#
and #x_2=(+31-sqrt((961+36)))/(2)=-0.287653#

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