How do you solve for x in the equation ${(x - 3)}^{2} + 25 = 0$ $(x-3)^2+25=0$ ?

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Answer 1 · 2018-03-16T01:49:05

Move $25$ $25$ to the other side:

${(x - 3)}^{2} = 25$ $(x-3)^2=25$

Take the square root of both sides to get rid of the square:

$\sqrt{{(x - 3)}^{2}} = \sqrt{- 25}$ $sqrt((x-3)^2)=sqrt(-25)$

$x - 3 = \pm 5 i$ $x-3=+-5i$

$x = 3 \pm 5 i$ $x=3+-5i$

How do you solve for x in the equation (x-3)^2+25=0(x−3)2+25=0(x-3)^2+25=0?