How do you solve #x^2+20x+104=0# by completing the square?

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Answer 1 · 2016-08-13T17:35:29

#x = -10 +- 2i#

Move the constant term to the RHS.

#x^2 + 20x = -104#

Add the square of half the coefficient of the #x# term to both sides:

#x^2 + 20x + color(red)(10^2) = -104 + color(red)(10^2)#

This becomes:

#(x+10)^2 = -104+100#

#(x+10)^2 = -4#

Take square roots of both sides.

#x+10 = +-sqrt(-4) = +-sqrt(4i^2) = +-2i#

#x = -10 +- 2i#