# How do you find the zeros, real and imaginary, of y= -7x^2-3x+2  using the quadratic formula?

Jan 26, 2018

About $- 0.79$ and $0.36$.

#### Explanation:

This quadratic is in standard form: $a {x}^{2} + b x + c$

So for $y = - 7 {x}^{2} - 3 x + 2$,
$a = - 7$,
$b = - 3$, and
$c = 2$.

Now we use the quadratic formula

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

and plug in our $a$, $b$, and $c$ values and simplify:

$x = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(- 7\right) \left(2\right)}}{2 \left(- 7\right)}$

$= \frac{3 \pm \sqrt{9 + 56}}{- 14}$

$= \frac{3 \pm \sqrt{65}}{- 14}$

which splits into the two solutions

$x = \frac{3 + \sqrt{65}}{- 14} \approx - 0.79$
and
$x = \frac{3 - \sqrt{65}}{- 14} \approx 0.36$