How do you solve $x^{2} - 4 x + 5 = 0$ $x^2-4x+5=0$ by completing the square?

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Answer 1 · 2018-05-15T13:19:20

No solutions.

$x^{2} - 4 x = - 5$ $x^2-4x=-5$

${(x - 2)}^{2} - 4 = - 5$ $(x-2)^2-4=-5$

${(x - 2)}^{2} = - 1$ $(x-2)^2=-1$

$(x - 2) = \pm \sqrt{- 1}$ $(x-2)=pmsqrt(-1)$

cannot do the square root of a negative number so no solutions

How do you solve x2−4x+5=0x^2-4x+5=0 by completing the square?