Question

Factor: 2z^2 + 9z+9=0
Quadratic formula. Please help!

Answer 1

First, we can identify the coefficients of the quadratic equation as follows:

`$a=\boxed{2}, \quad b=\boxed{9}, \quad c=\boxed{9}$`

The quadratic formula is:

`$z = (-b \pm √(b^2 - 4ac))/(2a)$`

We can substitute the values of a, b, and c into the formula:

`$z = \frac{-\boxed{9} \pm \sqrt{\boxed{9}^2 - 4(\boxed{2})(\boxed{9})}}{2(\boxed{2})}$`

Simplifying under the square root:

`$z = \frac{-\boxed{9} \pm \sqrt{\boxed{9}}}{\boxed{4}}$`

Simplifying further:

`$z = \frac{-\boxed{9} \pm \boxed{3}}{\boxed{4}}$`

So the two solutions are:

`$z_1 = \frac{-\boxed{9} + \boxed{3}}{\boxed{4}} = \boxed{-(3)/(2)}, \quad z_2 = \frac{-\boxed{9} - \boxed{3}}{\boxed{4}} = \boxed{-3}$`