How do you find the zeros, real and imaginary, of #y=-5x^2-2x+29# using the quadratic formula?

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Answer 1 · 2016-02-22T09:26:49

Zeros of #y=−5x^2−2x+29# are
#-1/5+sqrt146/5# and #-1/5+sqrt146/5#

To find the zeros, real and imaginary, of #y=−5x^2−2x+29#, we need to identify values of #x# that make #y=0# i.e. #−5x^2−2x+29=0# or #5x^2+2x-29=0#.

As solution of #ax^2+bx+x=0# is given by #(-b+-sqrt(b^2-4ac))/(2a)# and as in #5x^2+2x-29=0#, #a=5, b=2, c=-29#

Hence, zeros of #5x^2+2x-29=0# are given by

#(-2+-sqrt(2^2-4*5*(-29)))/(2*5)# or

#(-2+-sqrt(4+580))/10# or #(-2+-sqrt584)/10# or

#(-2+-sqrt584)/10# or #(-2+-2sqrt146)/10#

Hence, zeros of #y=−5x^2−2x+29# are
#-1/5+sqrt146/5# and #-1/5+sqrt146/5#