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

1 Answer
Mar 12, 2017

Answer:

#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#.