How do you solve #(x-3)^2+25=0#?

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Answer 1 · 2014-11-14T20:40:22

If you are looking for real solutions, then there are none.

If you are looking for complex solutions, then

#(x-3)^2+25=0#

by subtracting #25#,

#=> (x-3)^2=-25#

by taking the square-root,

#=>x-3=pmsqrt{-25}=pm5i#

by adding #3#,

#=> x=3 pm 5i#

I hope that this was helpful.