How do you find the zeros, real and imaginary, of #y=- x^2-22x+67 # using the quadratic formula?

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Answer 1 · 2016-04-28T09:32:14

Zeros of #y=-x^2-22x+47# are #-11+2sqrt67# and #-11-2sqrt47#

To find zeros of #y=-x^2-22x+67#, one needs to find values of #x# for which #y=f(x)=0#.

For #ax^2+bx+c=0#, solution is given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#. Hence for #-x^2-22x+67=0# or #x^2+22x-67=0#,

#x=(-22+-sqrt(22^2-4xx1xx(-67)))/2#

= #(-22+-sqrt(484+268))/2#

= #(-22+-sqrt(752))/2#

= #(-22+-4sqrt47)/2#

= #-11+-2sqrt47#