# How do you find the zeros, real and imaginary, of y=-4x^2-4x-15 using the quadratic formula?

Nov 17, 2015

Finding the values of $x$ for $\textcolor{w h i t e}{\times x} y = - 4 {x}^{2} - 4 x - 15 = 0$

$x \notin \mathbb{R} \textcolor{w h i t e}{\times}$ but $\textcolor{w h i t e}{\times} x \in \mathbb{C} \to x = - \frac{1}{2} \pm i \sqrt{\frac{7}{2}}$

#### Explanation:

Using standard for of $y = a {x}^{2} + b x + c$

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

Let:
$a = - 4$
$b = - 4$
$c = - 15$

So by substitution we have:

x= (-(-4) +- sqrt((-4)^2 -(4)(-4)(-15)))/(2(-4)

$x = \frac{+ 4 \pm \sqrt{16 - 240}}{2 \left(- 4\right)}$

$x = \frac{4 \pm \sqrt{- 224}}{- 8}$

partitioning 224 into prime numbers and squaring where able

$x = \frac{4 \pm \sqrt{\left(- 1\right) \times {2}^{2} \times {2}^{2} \times 2 \times 7}}{- 8}$

$x = \frac{4 \pm 4 \sqrt{- 14}}{- 8}$

$x = \frac{1 \pm i \sqrt{14}}{- 2}$
As we have $\pm$ in the numerator before the root having a negative denominator makes no tangible difference to that root. However, it does have an effect on the 1 preceding it. Consequently we have $- \frac{1}{2}$.

$x = - \frac{1}{2} \pm i \sqrt{\frac{14}{4}}$

$x = - \frac{1}{2} \pm i \sqrt{\frac{7}{2}}$