# How do you solve x^2+4x+2=0 using the quadratic formula?

Mar 4, 2018

See a solution process below:

#### Explanation:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

For: $1 {x}^{2} + 4 x + 2 = 0$

We can substitute:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{4}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{2}$ for $\textcolor{g r e e n}{c}$ giving:

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{{\textcolor{b l u e}{4}}^{2} - \left(4 \times \textcolor{red}{1} \times \textcolor{g r e e n}{2}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{16 - 8}}{2}$

$x = \frac{- \textcolor{b l u e}{4} - \sqrt{8}}{2}$ and $x = \frac{- \textcolor{b l u e}{4} + \sqrt{8}}{2}$

$x = - \frac{\textcolor{b l u e}{4}}{2} - \frac{\sqrt{8}}{2}$ and $x = - \frac{\textcolor{b l u e}{4}}{2} + \frac{\sqrt{8}}{2}$

$x = - 2 - \frac{\sqrt{8}}{2}$ and $x = - 2 + \frac{\sqrt{8}}{2}$

$x = - 2 - \frac{\sqrt{4 \times 2}}{2}$ and $x = - 2 + \frac{\sqrt{4 \times 2}}{2}$

$x = - 2 - \frac{\sqrt{4} \sqrt{2}}{2}$ and $x = - 2 + \frac{\sqrt{4} \sqrt{2}}{2}$

$x = - 2 - \frac{2 \sqrt{2}}{2}$ and $x = - 2 + \frac{2 \sqrt{2}}{2}$

$x = - 2 - \sqrt{2}$ and $x = - 2 + \sqrt{2}$