How do you solve the equation #2x^2+338=0#?

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Answer 1 · 2017-05-08T22:39:53

#x=+-13i#

First, we need to isolate #x#:

#2x^2+338=0#

subtract #338# on both sides

#2x^2=-338#

divide by #2#

#x^2=-169#

take the square root of both sides

#+-cancel(sqrt(x^cancel(2))##=+-sqrt(-169)#

#x=+-13i#

The #i# comes from #sqrt(-1)#