# How do you solve x²+6x+2=9?

Jun 11, 2018

$x = 1 \mathmr{and} x = - 7$

#### Explanation:

${x}^{2} + 6 x + 2 = 9$

Collecting like terms;

${x}^{2} + 6 x + 2 - 9 = 0$

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

Using Factorization Method..

Factors: $1 \mathmr{and} 7$

$6 x = 7 x - 1 x$

$- 7 = 7 \times - 1$

Therefore;

${x}^{2} - x + 7 x - 7 = 0$

Grouping;

$\left({x}^{2} - x\right) + \left(7 x - 7\right) = 0$

Factorizing;

$x \left(x - 1\right) + 7 \left(x - 1\right) = 0$

$\left(x - 1\right) \left(x + 7\right) = 0$

$x - 1 \mathmr{and} x + 7 = 0$

$x = 1 \mathmr{and} x = - 7$

Jun 11, 2018

$x = 1$ and $x = - 7$

#### Explanation:

We're dealing with a quadratic, so we want to set it equal to zero to find its zeroes. We now have

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

To factor this business, let's do a little thought experiment:

What two numbers sum up to $6$ and have a product of $- 7$? After some trial and error, we arrive at $- 1$ and $7$. Now, we can factor this as

$\left(x - 1\right) \left(x + 7\right) = 0$

Setting both factors equal to zero, we get

$x = 1$ and $x = - 7$

Hope this helps!