# How do you solve 9x^2 + 6x + 1= 0 using the quadratic formula?

Jun 6, 2017

See a solution process below:

#### Explanation:

For ${a}^{x} 2 + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substituting $9$ for $a$; $6$ for $b$ and $1$ for $c$ gives:

$x = \frac{- 6 \pm \sqrt{{6}^{2} - \left(4 \cdot 9 \cdot 1\right)}}{2 \cdot 9}$

$x = \frac{- 6 \pm \sqrt{36 - 36}}{2 \cdot 9}$

$x = \frac{- 6 \pm \sqrt{0}}{2 \cdot 9}$

$x = \frac{- 6 \pm 0}{2 \cdot 9}$

$x = - \frac{6}{18}$

$x = - \frac{1}{3}$