How do you find the zeros, real and imaginary, of #y=8x^2-4x-73# using the quadratic formula?

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Answer 1 · 2018-03-06T11:53:56

#x approx 3.281 or -2.781#

#y=8x^2-4x-73#

Let's first check the discriminant:
#Delta = (-4)^2 - 4xx8xx(-73)#

#= 16+2336 = 2352#

Since #Delta >0# #y# will have 2 real roots.

Apply the quadratic fomrula:

#x= (4+-sqrt(2352))/(2xx8)#

#approx (4+-48.4874)/16#

#approx 3.281 or -2.781#