How do you find the zeros, real and imaginary, of #y=x^2+32x+44# using the quadratic formula?

1 Answer
George C.
Dec 24, 2015

Substitute the coefficients into the quadratic formula to find:

#x=-16+-2sqrt(53)#

Explanation:

#x^2+32x+44# is of the form #ax^2+bx+c#
with #a=1#, #b=32# and #c=44#.

This has zeros given by the quadratic formula:

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

#=(-32+-sqrt(32^2-(4xx1xx44)))/(2*1)#

#=(-32+-sqrt(1024-176))/2#

#=(-32+-sqrt(848))/2#

#=(-32+-sqrt(4^2*53))/2#

#=(-32+-4sqrt(53))/2#

#=-16+-2sqrt(53)#

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