How do you find the zeros, real and imaginary, of # y=3x^2-7x-5 # using the quadratic formula?

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Answer 1 · 2016-09-14T13:26:43

Zeros are real and are #(7-sqrt109)/6# and #(7+sqrt109)/6#

The zeros of #y=ax^2+bx+c# are given by quadratic formula

and are #(-b+-sqrt(b^2-4ac))/(2a)#

Hence, zeros of #y=3x^2-7x-5# are given by

#(-(-7)+-sqrt((-7)^2-4xx3xx(-5)))/(2xx3)#

= #(7+-sqrt(49+60))/6#

= #(7+-sqrt109)/6#

Hence zeros are real and are #(7-sqrt109)/6# and #(7+sqrt109)/6#