How do you find the zeros, real and imaginary, of #y= 23x^2+18x-24# using the quadratic formula?

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Answer 1 · 2018-03-27T13:40:32

There are two real roots:

# -9/23-sqrt(633)/23 # and # -9/23+sqrt(633)/23 #

We have:

# y = 23x^2+18x-24 #

So the roots are given by the solution of the equation:

# 23x^2+18x-24 = 0#

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

With #a=23, b=18# and #c=-24# we have:

# x = (-18+-sqrt(18^2-4(23)(-24)))/(2(23)) #
# \ \ = (-18+-sqrt(324+2208))/(46) #
# \ \ = (-18)/(46)+-sqrt(2532)/(46) #
# \ \ = -9/23+-sqrt(633)/23 #

Hence, there are two real roots:

# -9/23-sqrt(633)/23 # and # -9/23+sqrt(633)/23 #