How do you find the zeros of # y = 3/2x^2 + 3/2x +9/2 # using the quadratic formula?

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Answer 1 · 2018-03-18T09:56:54

#x=(-1+-isqrt(11))/2#

Finding the zeroes of the function is the same as solving the following equation:

#3/2x^2+3/2x+9/2=0#

Because fractions are quite annoying to deal with, I will multiply both sides by #2 \/ 3# before we use the quadratic formula:

#2/3(3/2x^2+3/2x+9/2)=0*2/3#

#x^2+x+3=0#

Now we can use the quadratic formula, which says that if we have a quadratic equation in the form:

#ax^2+bx+c=0#

The solutions will be:

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

In this case, we get:

#x=(-1+-sqrt((-1)^2-4*3))/2#

#x=(-1+-sqrt(1-12))/2#

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

#x=(-1+-isqrt(11))/2#