How do you find the zeros, real and imaginary, of #y=-5x^2-2x-9# using the quadratic formula?

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Answer 1 · 2017-06-09T02:57:37

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#; #-2# for #b# and #-9# for #c# gives:

#x = (-(-2) +- sqrt((-2)^2 - (4 * -5 * -9)))/(2 * -5)#

#x = (2 +- sqrt(4 - 180))/(-10)#

#x = (2 +- sqrt(-176))/(-10)#

#x = (2 +- sqrt(16 * -11))/(-10)#

#x = (2 +- (sqrt(16) * sqrt(-11)))/(-10)#

#x = (2 +- 4sqrt(-11))/(-10)#

#x = -(1 +- 2sqrt(-11))/5#