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

Mar 18, 2018

$x = \frac{- 1 \pm i \sqrt{11}}{2}$

#### Explanation:

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

$\frac{3}{2} {x}^{2} + \frac{3}{2} x + \frac{9}{2} = 0$

Because fractions are quite annoying to deal with, I will multiply both sides by $2 \frac{\setminus}{3}$ before we use the quadratic formula:

$\frac{2}{3} \left(\frac{3}{2} {x}^{2} + \frac{3}{2} x + \frac{9}{2}\right) = 0 \cdot \frac{2}{3}$

${x}^{2} + x + 3 = 0$

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

$a {x}^{2} + b x + c = 0$

The solutions will be:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case, we get:

$x = \frac{- 1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 3}}{2}$

$x = \frac{- 1 \pm \sqrt{1 - 12}}{2}$

$x = \frac{- 1 \pm \sqrt{- 11}}{2}$

$x = \frac{- 1 \pm i \sqrt{11}}{2}$