How do you find the zeros, real and imaginary, of #y= 8x^2+3x-2# using the quadratic formula?

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Answer 1 · 2016-12-28T14:14:21

We have real zeros #(-3+sqrt73)/16# and #(-3-sqrt73)/16#

Quadratic formula gives the zeros of a quadratic function #y=ax^2+bx+c# as #(-b+-sqrt(b^2-4ac))/(2a)#.

Hence, zeros of a function #y=8x^2+3x-2# are

#(-3+-sqrt(3^2-4xx8xx(-2)))/(2xx8)#

or #(-3+-sqrt(9+64))/16#

or #(-3+-sqrt73)/16#

i.e. #(-3+sqrt73)/16# and #(-3-sqrt73)/16#

i.e. we have real zeros.