# How do you find the zeros, real and imaginary, of y= -3x^2+3x-18  using the quadratic formula?

Feb 4, 2016

The two imaginary roots of this equation are $= \frac{1}{2} - \frac{\sqrt{207} i}{6} \mathmr{and} \frac{1}{2} + \frac{\sqrt{207} i}{6.}$

#### Explanation:

For a quadratic equation in standard form, $a {x}^{2} + b x + c = 0$, the quadratic formula is a means of finding the roots ('zeros') of the equation - the points where the quadratic (which will be parabola-shaped) crosses the x-axis, which is the line $y = 0$.

We can take the equation we are given and set $y = 0$ and it will be in standard form.

The quadratic formula has the form:

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

If the piece under the square root sign is positive, there will be two real roots. If it is negative there may be imaginary roots.

In this case:

$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot \left(- 3\right) \cdot \left(- 18\right)}}{2 \left(- 3\right)} = \frac{- 3 \pm \sqrt{9 - 216}}{-} 6$

$x = \frac{- 3 \pm \sqrt{- 207}}{-} 6 = \frac{- 3 \pm \sqrt{207} \sqrt{- 1}}{-} 6 = \frac{- 3 \pm \sqrt{207} i}{-} 6$

$x = \frac{1}{2} - \frac{\sqrt{207} i}{6} \mathmr{and} \frac{1}{2} + \frac{\sqrt{207} i}{6.}$

In this case both roots are imaginary.