# How do you solve for all solutions, including complex solutions, to (x + 14)^2 = 21 ?

Mar 12, 2017

$x = \setminus \pm \sqrt{21} - 14$
There are no complex solutions.

#### Explanation:

By squaring a monomial, we will get a quadratic. Thus, we have two solutions to the equation.

By taking the square root of both sides, we find that
$x + 14 = \setminus \pm \sqrt{21}$, because taking the square root can result in either a plus or a minus solution (squaring a real number will always come out positive).
Then, our answer comes after subtracting $14$ on both sides.
$x = \setminus \pm \sqrt{21} - 14$.