# How do you solve the equation (3x+2)^2-49=0?

Feb 19, 2017

$x = \frac{5}{3}$ and $x = - 3$

#### Explanation:

${\left(3 x + 2\right)}^{2} - 49 = 0$ can be written as

${\left(3 x + 2\right)}^{2} - {7}^{2} = 0$

As ${a}^{2} - {b}^{2}$ can be factorized as $\left(a + b\right) \left(a - b\right)$, this can be factorized as

$\left(3 x + 2 + 7\right) \left(3 x + 2 - 7\right) = 0$

Hence, either $3 x + 2 + 7 = 0$ i.e. $3 x = - 9$ and $x = - 3$

or $3 x + 2 - 7 = 0$ i.e. $3 x = 7 - 2 = 5$ and $x = \frac{5}{3}$

Feb 19, 2017

Alternative to get $x = \frac{5}{3}$ and $x = - 3$

#### Explanation:

You can also treat this as a completing the square problem where the step of making the square is already done for you.

Move the 49 to the other side:

${\left(3 x + 2\right)}^{2} = 49$

Square root both sides:

$3 x + 2 = \left(\frac{+}{-}\right) 7$

Treat it as two problems and solve for x

$3 x + 2 = 7$

$3 x = 5$

$x = \frac{5}{3}$

$3 x + 2 = - 7$
$3 x = - 9$
$x = - 3$