How do you find the zeros, real and imaginary, of #y= -x^2-55x-8 # using the quadratic formula?

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Answer 1 · 2017-07-09T02:58:51

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 #-1# for #a#; #-55# for #b# and #-8# for #c# gives:

#x = (-(-55) +- sqrt((-55)^2 - (4 * -1 * -8)))/(2* -1)#

#x = (55 +- sqrt(3025 - 31))/-2#

#x = (55 +- sqrt(2994))/-2#

#x = -55/2 + sqrt(2994)/2# and #x = -55/2 - sqrt(2994)/2#