Question

hw 7.3 (e - book 3.6) solving quadratic equations. question 3 of 8 (1 point) | question attempt: 1 of unlimited. a = 1, b = - 3, c = - 9. part: 1 / 4. part 2 of 4. apply the quadratic formula. y = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. y = \frac{-(-3)\pm\sqrt{(\square)^{2}-4(1)(\square)}}{2(\square)}

Explanation:

Step1: Substitute values of a, b, c

Substitute $a = 1$, $b=-3$, $c = - 9$ into the quadratic formula $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
$y=\frac{-(-3)\pm\sqrt{(-3)^{2}-4(1)(-9)}}{2(1)}$

Answer:

$y=\frac{-(-3)\pm\sqrt{(-3)^{2}-4(1)(-9)}}{2(1)}$