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

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Answer 1 · 2017-08-08T02:21:03

#x==1+sqrt3i#
#x=-1-sqrt3i#

Identify the values of #a,b & c# and substitute in using the quadratic formula: # x = (-b \pm sqrt(b^2-4ac)) / (2a) #

#a=-1#

#b=-2#

#c=-4#

# x = (-(-2) \pm sqrt((-2)^2-4(-1)(-4))) / (2(-1)) #

# x = (2 \pm sqrt(4-16)) / (-2) #

# x = (2 \pm sqrt(-12)) / (-2) #

# x = (2 \pm 2 sqrt(3)i) / (-2) #

# x = -1 \pm sqrt(3)i#

#x=1+sqrt3i#, #x=-1-sqrt3ilarr# Final solutions