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Solve by Completing the square. b^(2)+2b+10=0
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b= -1 plus or minus 3i
We know \[(a+b)^2=a^2+2ab+b^2\] here we have \[b^2+2b+10=0\] \[b^2+2b+1+9=0\] \[(b+1)^2+9=0\] or \[(b+1)^2=-9\] we get \[(b+1)=\pm 3i \] \[( i =\sqrt{-1} )\] we get \[b=-1\pm3i\]
b^2+2b+10=0 b^2 + 1 = -10 +1 ( b + 1)^2 = -9 -> b+1 = +/- sqrt ( -9) = +/-3i => b = -1 +/-3i
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