How do you find the zeros, real and imaginary, of #y= -9x^2-28x-73# using the quadratic formula?

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Answer 1 · 2016-04-28T07:08:11

Zeros of #y=-9x^2-28x-73# are #-14/9-sqrt(461)/9i# and #-14/9+sqrt(461)/9i#

To find zeros of #y=-9x^2-28x-73#, 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 #-9x^2-28x-73=0# or #9x^2+28x+73=0#,

#x=(-28+-sqrt(28^2-4xx9xx73))/(2xx9)#

= #(-28+-sqrt(784-2628))/18#

= #(-28+-sqrt(-1844))/18#

= #(-28+-sqrt(-4xx461))/18#

= #-14/9+-sqrt(461)/9i#