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

Aug 8, 2017

The solutions of this equation are imaginary.

The roots of this equation are:

$x = - 1.07 i , x = 0.548 i$

#### Explanation:

The zeroes (solutions) occur when $y = 0$, so we have −9/2x^2+3/2x–3/2=0.

Standard form for a quadratic equation is $a {x}^{2} + b x + c = 0$, so in this case, $a = - \frac{9}{2}$, $b = \frac{3}{2}$ and $c = - \frac{3}{2}$.

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

Substituting in the values of $a , b$ and $c$, we have:

$x = \frac{- \frac{3}{2} \pm \sqrt{{\left(\frac{3}{2}\right)}^{2} - 4 \times \left(- \frac{9}{2}\right) \times \left(- \frac{3}{2}\right)}}{2 \left(\frac{3}{2}\right)}$
$= \frac{- \frac{3}{2} \pm \sqrt{\left(\frac{9}{4} - 27\right)}}{6}$

Since the total under the square root sign yields a negative number, there are no real roots of the equation, only imaginary roots.

$\sqrt{\frac{9}{4} - 27} = 4.97 i$ where $i = \sqrt{- 1}$.

That means the roots of this equation are:

$x = - 1.07 i , x = 0.548 i$