How do you find the zeros, real and imaginary, of #y= -3x^2-8x-18 #using the quadratic formula?

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Answer 1 · 2016-06-09T04:16:24

Zeros of #y=-3x^2-8x-18# are

#-4/3-2sqrt38i# and #-4/3+2sqrt38i#

Zeros of #y=ax^2+bx+c# are given by the quadratic formula as

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

Hence for #y=-3x^2-8x-18# these are given by

#(-(-8)+-sqrt((-8)^2-4xx(-3)xx(-18)))/(2*(-3))=(8+-sqrt(64-216))/(-6)#

= #-4/3+-sqrt(-152)/6=-4/3+-i2sqrt38#

Hence zeros of #y=-3x^2-8x-18# are

#-4/3-2sqrt38i# and #-4/3+2sqrt38i#