Question

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

Answer 1

`x=\pmsqrt(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=\pmsqrt(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=\pmsqrt(21)-14`.