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

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

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Answer 1 · 2017-03-15T11:53:21

No real roots

Explanation:

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

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

1 Answer

Explanation:

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