# How do I solve this quadratic equation?

## $6 {x}^{2} + 7 x + 2 = 0$

Mar 9, 2018

$x = - \frac{1}{2}$ and $x = - \frac{2}{3}$

#### Explanation:

$6 {x}^{2} + 7 x + 2$

can be factored into a binomial,

$\left(3 x + \frac{3}{2}\right) \left(2 x + \frac{4}{3}\right)$

By setting a factor to zero we can solve for an x value
$3 x + \frac{3}{2} = 0$
$x = - \frac{1}{2}$

$2 x + \frac{4}{3} = 0$
$x = - \frac{2}{3}$

Mar 9, 2018

$x = - \frac{1}{2} , - \frac{2}{3}$

#### Explanation:

We can solve this quadratic with the strategy factoring by grouping. Here, we will rewrite the $x$ term as the sum of two terms, so we can split them up and factor. Here's what I mean:

$6 {x}^{2} + \textcolor{b l u e}{7 x} + 2 = 0$

This is equivalent to the following:

$6 {x}^{2} + \textcolor{b l u e}{3 x + 4 x} + 2 = 0$

Notice, I only rewrote $7 x$ as the sum of $3 x$ and $4 x$ so we can factor. You'll see why this is useful:

$\textcolor{red}{6 {x}^{2} + 3 x} + \textcolor{\mathmr{and} a n \ge}{4 x + 2} = 0$

We can factor a $3 x$ out of the red expression, and a $2$ out of the orange expression. We get:

$\textcolor{red}{3 x \left(2 x + 1\right)} + \textcolor{\mathmr{and} a n \ge}{2 \left(2 x + 1\right)} = 0$

Since $3 x$ and $2$ are being multiplied by the same term ($2 x + 1$), we can rewrite this equation as:

$\left(3 x + 2\right) \left(2 x + 1\right) = 0$

We now set both factors equal to zero to get:

$3 x + 2 = 0$

$\implies 3 x = - 2$

$\textcolor{b l u e}{\implies x = - \frac{2}{3}}$

$2 x + 1 = 0$

$\implies 2 x = - 1$

$\textcolor{b l u e}{\implies x = - \frac{1}{2}}$

Our factors are in blue. Hope this helps!

Mar 9, 2018

$- \frac{1}{2} = x = - \frac{2}{3}$

#### Explanation:

Hmm...
We have:
$6 {x}^{2} + 7 x + 2 = 0$ Since ${x}^{2}$ is being multiplied by a number here, let's multiply $a$ and $c$ in $a {x}^{2} + b x + c = 0$

$a \cdot c = 6 \cdot 2 \implies 12$

We ask ourselves: Do any of the factors of $12$ add up to $7$?

Let's see...

$1 \cdot 12$ Nope.

$2 \cdot 6$ Nope.

$3 \cdot 4$ Yep.

We now rewrite the equation like the following:

$6 {x}^{2} + 3 x + 4 x + 2 = 0$ (The order of $3 x$ and $4 x$ does not matter.)

Let's separate the terms like this:

$\left(6 {x}^{2} + 3 x\right) + \left(4 x + 2\right) = 0$ Factor each parenthesis.

$\implies 3 x \left(2 x + 1\right) + 2 \left(2 x + 1\right) = 0$

For better understanding, we let $n = 2 x + 1$

Replace $2 x + 1$ with $n$.

$\implies 3 x n + 2 n = 0$ Now, we see that each group have $n$ in common.

Let's factor each term.

$\implies n \left(3 x + 2\right) = 0$ Replace $n$ with $2 x + 1$

$\implies \left(2 x + 1\right) \left(3 x + 2\right) = 0$

Either $2 x + 1 = 0$ or $3 x + 2 = 0$

Let's solve each case.

$2 x + 1 = 0$

$2 x = - 1$

$x = - \frac{1}{2}$ That's one answer.

$3 x + 2 = 0$

$3 x = - 2$

$x = - \frac{2}{3}$ That's another.