# How do you find the zeros, if any, of y= -17x^2+6x-3using the quadratic formula?

Mar 15, 2017

No real roots

#### Explanation:

$- b \pm \frac{\sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$
$a = - 17$
$b = 6$
$c = - 3$
substitute
$- \left(6\right) \pm \frac{\sqrt{{\left(6\right)}^{2} - 4 \left(- 17\right) \left(- 3\right)}}{2 \left(- 17\right)}$
since ${6}^{2} - 4 \left(- 17\right) \left(- 3\right)$ is negative, you cannot take the square root of it and get real numbers. Therefore there are no real roots