# How do you find the zeros, real and imaginary, of y= -7x^2+4x+2  using the quadratic formula?

May 24, 2017

$x = \frac{2}{7} + \frac{3}{7} \sqrt{2} \text{ }$ or $\text{ } x = \frac{2}{7} - \frac{3}{7} \sqrt{2}$

#### Explanation:

$y = - 7 {x}^{2} + 4 x + 2$

is in the form:

$y = a {x}^{2} + b x + c$

with $a = - 7$, $b = 4$ and $c = 2$

This has zeros given by the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- \textcolor{b l u e}{4} \pm \sqrt{{\textcolor{b l u e}{4}}^{2} - 4 \left(\textcolor{b l u e}{- 7}\right) \left(\textcolor{b l u e}{2}\right)}}{2 \left(\textcolor{b l u e}{- 7}\right)}$

$\textcolor{w h i t e}{x} = \frac{- 4 \pm \sqrt{16 + 56}}{- 14}$

$\textcolor{w h i t e}{x} = \frac{- 4 \pm \sqrt{72}}{- 14}$

$\textcolor{w h i t e}{x} = \frac{- 4 \pm \sqrt{{6}^{2} \cdot 2}}{- 14}$

$\textcolor{w h i t e}{x} = \frac{- 4 \pm 6 \sqrt{2}}{- 14}$

$\textcolor{w h i t e}{x} = \frac{- 2 \pm 3 \sqrt{2}}{- 7}$

$\textcolor{w h i t e}{x} = \frac{2}{7} \pm \frac{3}{7} \sqrt{2}$

That is:

$x = \frac{2}{7} + \frac{3}{7} \sqrt{2} \text{ }$ or $\text{ } x = \frac{2}{7} - \frac{3}{7} \sqrt{2}$