# How do you find the zeros, real and imaginary, of y=- 7x^2-15x -35 using the quadratic formula?

Jul 8, 2017

$x = \frac{15 + i \sqrt{755}}{- 14} ,$$\frac{15 - i \sqrt{755}}{- 14}$

#### Explanation:

The zeros of a quadratic equation are the values for $x$, and are where the parabola crosses the x-axis.

$y = - 7 {x}^{2} - 15 x - 35$

Substitute a zero for the $y$.

$0 = - 7 {x}^{2} - 15 x - 35$ is a quadratic equation in the form: $a {x}^{2} + b x + c = 0$, where $a = - 7$, $b = - 15$, $c = - 35$.

Use the quadratic formula to solve for $x$ (the zeros).

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

Substitute the given values for $a , b , \mathmr{and} c$ into the formula.

$x = \frac{- \left(- 15\right) \pm \sqrt{{\left(- 15\right)}^{2} - 4 \cdot - 7 \cdot - 35}}{2 \cdot - 7}$

Simplify.

$x = \frac{15 \pm \sqrt{- 755}}{- 14}$

$x = \frac{15 + i \sqrt{755}}{- 14} ,$$\frac{15 - i \sqrt{755}}{- 14}$

Jul 8, 2017

2 imaginary roots:
$x = \frac{- 15 \pm \sqrt{755}}{14}$

#### Explanation:

$y = - \left(7 {x}^{2} + 15 x + 35\right) = 0$
Use new improved quadratic formula (Socratic Search)
$D = {d}^{2} = {b}^{2} - 4 a c = 225 - 980 = - 755 = 755 {i}^{2}$ -->
$d = \pm i \sqrt{755}$
Because D < 0, there are 2 imaginary roots.
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{15}{14} \pm i \frac{\sqrt{755}}{14} = \frac{- 15 \pm i \sqrt{755}}{14}$

Note. Using the improved quadratic formula, the proceeding is much simpler, and the numeric computation is much easier.