# How do you find the zeros, real and imaginary, of y=-4x^2-2x+33 using the quadratic formula?

Apr 16, 2017

$x \approx - 3.133$ and $\approx 2.633$

#### Explanation:

The quadratic formula is bases on $a {x}^{2} + b x + c$:
$\frac{- b \pm \sqrt{{b}^{2} - \left(4 \cdot a \cdot c\right)}}{2 \cdot a}$

So, our equation of $y = - 4 {x}^{2} - 2 x + 33$ tells us $a = - 4$, $b = - 2$, $c = 33$

Thus: $\frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - \left(4 \cdot - 4 \cdot 33\right)}}{2 \cdot - 4}$

$\frac{2 \pm \sqrt{4 - \left(4 \cdot - 4 \cdot 33\right)}}{- 8}$

$\frac{2 \pm \sqrt{4 - - 528}}{-} 8$

$\frac{2 \pm \sqrt{532}}{- 8}$

$\frac{2 \pm 2 \sqrt{133}}{- 8}$

$\frac{2 \left(1 \pm \sqrt{133}\right)}{- 8}$

$\frac{1 \pm \sqrt{133}}{-} 4$

That's the exact form, but the estimated form is $x \approx - 3.133$ and $\approx 2.633$