# How do you solve x^2 +6x +8 =0 using the quadratic formula?

May 8, 2018

The answers are $x = - 2$ and $x = - 4$.

#### Explanation:

To start, the quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this problem, $a = 1$ (as the ${x}^{2}$ term has no coefficient), $b = 6$, and $c = 8$.

Plug those values into the quadratic equation to get:

$x = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \left(1\right) \left(8\right)}}{2 \left(1\right)}$

Multiply $2 \cdot 1$ on the bottom of the fraction:

$x = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \left(1\right) \left(8\right)}}{2}$

Square $6$ and multiply $4 \cdot 1 \cdot 8$ within the square root:

$x = \frac{- 6 \pm \sqrt{36 - 32}}{2}$

Subtract $36 - 32$ inside the root:

$x = \frac{- 6 \pm \sqrt{4}}{2}$

Solve for $\sqrt{4}$

$x = \frac{- 6 \pm 2}{2}$

If the $\pm$ is positive, you get

$x = \frac{- 6 + 2}{2}$, which simplifies to $x = \frac{- 4}{2}$, or $\textcolor{red}{- 2}$

If the $\pm$ is negative, you get

$x = \frac{- 6 - 2}{2}$, which simplifies to $x = \frac{- 8}{2}$, or $\textcolor{red}{- 4}$

May 8, 2018

x = -2 or x = -4

#### Explanation:

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

You have...
$a {x}^{2} + b x + c = 0$
${x}^{2} + 6 x + 8 = 0$

With that you can do...
${x}^{2} + 6 x + 8 = 0$
$a = 1$
$b = 6$
$c = 8$

Then you substitute what you have into the quadratic formula
When you do that you get...
$x = \frac{- \left(6\right) \pm \sqrt{{\left(6\right)}^{2} - 4 \left(1\right) \left(8\right)}}{2 \left(a\right)}$

The '$\pm$' means that your going to have 2 answers like x = this or that so, you can just solve the whole equation using '$+$' first then use '$-$' next.

when you do that you'll get an answer of $x = - 2 \mathmr{and} - 4$