# How do you find the zeroes for y=9x^2+6x-1?

Feb 13, 2017

-0.805 or 0.138

#### Explanation:

Assuming $y = 0$ you can use the quadratic formula

x=(-b±sqrt(b^2-4ac))/(2a)

to find the roots of a quadratic.
A quadratic is defined as $a {x}^{2} + b x + c$ so for your case you may plug in the respective values of $a$, $b$, and $c$:

$a = 9$
$b = 6$
$c = - 1$

Then the quadratic formula would look like this:

x=(-6±sqrt(6^2-4*9*-1))/(2*9)

Due to the ± sign we then have two variants for 2 answers (2 is also the degree of the polynomial which determines the amount of roots/answers of the polynomial)

$x = \frac{- 6 + \sqrt{{6}^{2} - 4 \cdot 9 \cdot - 1}}{2 \cdot 9} = 0.138071187458$
$x = \frac{- 6 - \sqrt{{6}^{2} - 4 \cdot 9 \cdot - 1}}{2 \cdot 9} = - 0.804737854124$

I rounded the answers to $x = - 0.805 , x = 0.138$