# How do you find the zeros of  y = 4.5x^2 – 1.5x + 1.5 using the quadratic formula?

Jun 14, 2017

$y$ has no real zeroes, but its complex zeroes are $\frac{3 \pm i \sqrt{87}}{16}$

#### Explanation:

The first thing we should check when trying to find the zeroes of a function is if its discriminant is greater than, equal to or less than zero.

${\Delta}_{y} = {1.5}^{2} - 4 \left(4\right) \left(1.5\right) = - \frac{87}{4}$

${\Delta}_{y} < 0 \therefore y$ has no real roots.

However, we can still find its complex zeroes using the quadratic formula.

$x = \frac{1.5 \pm \sqrt{- {1.5}^{2} - 4 \left(4\right) \left(1.5\right)}}{2 \left(4\right)} = \frac{3 \pm i \sqrt{87}}{16}$