How do you find the zeros, real and imaginary, of #y=-5x^2-13x-4# using the quadratic formula?

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Answer 1 · 2017-07-09T02:15:56

See a solution process below:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Substituting #-5# for #a#; #-13# for #b# and #-4# for #c# gives:

#x = (-(-13) +- sqrt((-13)^2 - (4 * -5 * -4)))/(2 * -5)#

#x = (13 +- sqrt(169 - 80))/(-10)#

#x = (13 +- sqrt(89))/(-10)#

#x = -13/10 +- -sqrt(89)/10#

#x = -13/10 + sqrt(89)/10# and #x = -13/10 - sqrt(89)/10#