# How do you find the zeros, real and imaginary, of y=-4x^2-2x+3 using the quadratic formula?

Feb 19, 2016

$\frac{- 1 \pm - 1 \sqrt{13}}{4}$

#### Explanation:

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

Let's look at our equation, $- 4 {x}^{2} - 2 x + 3$, in a more general form: $a {x}^{2} - b x + c$.

So $a$ is $- 4$
$b$ is $- 2$
$c$ is $3$.

Now let's plug our variables into the quadratic formula. I like to put parentheses around the variables I'm replacing, just to be extra careful about silly little sign errors.

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

This becomes $\frac{2}{-} 8 \pm \frac{\sqrt{4 - \left(- 48\right)}}{-} 8$, which we can simplify to $- \frac{1}{4} \pm \frac{2 \sqrt{13}}{-} 8$, or $- \frac{1}{4} \pm \frac{- 1 \sqrt{13}}{4}$.

The final answer is $\frac{- 1 \pm - 1 \sqrt{13}}{4}$