How do you solve $x^{2} - 3 x = - 10$ $x^2-3x=-10$ by completing the square?

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Answer 1 · 2015-05-28T19:54:46

$x^{2} - 3 x = - 10$ $x^2-3x=-10$

$x^{2} - 4 x + 4 = - x - 6$ $x^2-4x+4=-x-6$

${(x - 2)}^{2} = - x - 6$ $(x-2)^2=-x-6$

$x - 2 = \sqrt{- x - 6}$ $x-2=sqrt(-x-6)$

No real solutions, because you can't have the square root of a negative number.

How do you solve x2−3x=−10x^2-3x=-10 by completing the square?