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:

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

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

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

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

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