# How do you find the zeros, if any, of y= 2x^2 +x- 2using the quadratic formula?

Dec 24, 2015

$x = \frac{- 1 \pm \sqrt{17}}{4}$

#### Explanation:

x=(-b+-sqrt(b^2-4ac))/(2a

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

$y = 2 {x}^{2} + x - 2$,

$a = 2$
$b = 1$
$c = - 2$

Plug these into the quadratic formula:

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

$x = \frac{- 1 \pm \sqrt{1 - \left(- 16\right)}}{4}$

$x = \frac{- 1 \pm \sqrt{17}}{4}$

These are the two times when the function is equal to $0$ (the roots of the function).