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

Jan 23, 2016

There are two real zeroes:
$\textcolor{w h i t e}{\text{XXX")x=-4/7+sqrt(22)/7color(white)("XX}}$ and $\textcolor{w h i t e}{\text{XX}} x = - \frac{4}{7} - \frac{\sqrt{22}}{7}$

#### Explanation:

The quadratic formula for an expression of the form
$\textcolor{w h i t e}{\text{XXX}} a {x}^{2} + b x + c$
tells us that the zeroes occur at
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For the given expression: $- 7 {x}^{2} - 2 x + 3$
$\textcolor{w h i t e}{\text{XXX}} a = - 7$
$\textcolor{w h i t e}{\text{XXX}} b = - 2$
$\textcolor{w h i t e}{\text{XXX}} c = + 3$

So the zeroes are at
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} + 4 \left(- 7\right) \left(3\right)}}{2 \left(- 7\right)}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{4 \pm \sqrt{4 + 84}}{- 14}$
$\textcolor{w h i t e}{\text{XXX}} = - \frac{4 \pm 2 \sqrt{22}}{14}$
$\textcolor{w h i t e}{\text{XXX}} = - \frac{2}{7} \pm \frac{\sqrt{22}}{7}$

Note that since the argument of the square root is positive,
the zeroes are Real.