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

Dec 13, 2017

$x = \frac{3 + \sqrt{13}}{2}$ or $x = \frac{3 - \sqrt{13}}{2}$.

#### Explanation:

Start with $y = \frac{3}{2} {x}^{2} - \frac{9}{2} x - \frac{3}{2}$.

To find the zeros set $y$ equal to 0:

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

Since the equation equals 0 we can multiply through by 2 to clear the denominators:

$3 {x}^{2} - 9 x - 3 = 0$

Now notice we can divide by 3 to further simplify:

${x}^{2} - 3 x - 1 = 0$

Now we have a quadratic equation with $a = 1$, $b = - 3$, and $c = - 1$ and the quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

So we have:

$x = \frac{- \left(- 3\right) + \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 1\right)}}{2 \left(1\right)} = \frac{3 + \sqrt{13}}{2}$

or we have:

$x = \frac{- \left(- 3\right) - \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 1\right)}}{2 \left(1\right)} = \frac{3 - \sqrt{13}}{2}$

So, $x = \frac{3 + \sqrt{13}}{2}$ or $x = \frac{3 - \sqrt{13}}{2}$.