How do you find the zeros, real and imaginary, of #y=- 5x^2-15x -3# using the quadratic formula?

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Answer 1 · 2018-08-04T07:03:42

#x = (15 + sqrt(165))/-10# and #x = (15 - sqrt(165))/-10#

#y = -5x^2 - 15x - 3#

The quadratic formula is #x = (-b +- sqrt(b^2 - 4ac))/(2a)#.

We know that: #a = -5#, #b = -15#, and #c = -3# based on the equation, so let's plug them into the formula:

#x = (-(-15) +- sqrt((-15)^2 - 4(-5)(-3)))/(2(-5))#

#x = (15 +- sqrt(225 - 60))/-10#

#x = (15 +- sqrt(165))/-10#

Hope this helps!