# How do you solve x^2 + 9x + 9 = 0 using the quadratic formula?

Apr 26, 2016

Root1=$- 1.146$ and Root2 = $- 7.854$

#### Explanation:

This is quadratic equation where  a=1 ;b=9; c=9; b^2-4ac > 0 So roots are real. Roots are $- \frac{b}{2 \cdot a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$ $\therefore$Root1 $= - \frac{9}{2} + \frac{\sqrt{45}}{2} = - 1.146$ and Root2$= - \frac{9}{2} - \frac{\sqrt{45}}{2} = - 7.854$[Ans]

Apr 26, 2016

$\frac{- 9 \pm 3 \sqrt{5}}{2}$

#### Explanation:

$y = {x}^{2} + 9 x + 9 = 0$
Use the improved quadratic formula (Socratic Search)
$D = {d}^{2} = {b}^{2} - 4 a c = 81 - 36 = 45$--> $d = \pm 3 \sqrt{5}$
There are 2 real roots:

$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{9}{2} \pm 3 \frac{\sqrt{5}}{2} = \frac{- 9 \pm 3 \sqrt{5}}{2}$