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

Nov 14, 2014

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

If you are looking for complex solutions, then

${\left(x - 3\right)}^{2} + 25 = 0$

by subtracting $25$,

$\implies {\left(x - 3\right)}^{2} = - 25$

by taking the square-root,

$\implies x - 3 = \pm \sqrt{- 25} = \pm 5 i$

by adding $3$,

$\implies x = 3 \pm 5 i$

I hope that this was helpful.